maximus-21/PhysicsQA · Datasets at Hugging Face

question stringlengths 29 1.32k	input stringlengths 0 439	answer stringlengths 1 366	solution stringlengths 0 1.08k	topic stringclasses 29 values
Q4: In an interference experiment the ratio of amplitudes of coherent waves is (a1 / a2) = (1 / 3). The ratio of maximum and minimum intensities of fringes will be	(a) 4 (b) 9 (c) 2 (d) 18	(a) 4	(a1 / a2) = (1 / 3) Imax = a1 + a2 Imin = a1 - a2 (Imax / Imin) = [(a1 + a2)² / (a1 - a2)²] = [(1 + 1/3)² / (1 - 1/3)²] = (4/2)² = 4	Wave Optics
Q5: Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.	(a) 610 x 10^{-9} radian (b) 152.5 x 10^{-9} radian (c) 457.5 x 10^{-9} radian (d) 305 x 10^{-9} radian	(d) 305 x 10^{-9} radian	The limit of resolution, θΔ = 1.22λ/a = (1.22 x 500 x 10^{-9}) / (200 x 10^{-2}) = 3.05 x 10^{-7} radian = 305 x 10^{-9} radian	Wave Optics
Q6: In Young's double-slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be	(a) (√3 + 1)² : 16 (b) 9 : 1 (c) 25 : 9 (d) 4 : 1	(b) 9 : 1	I1 = 4I0 I2 = I0 Imax = (√I1 + √I2)² = (2√I0 + √I0)² = 9I0 Imin = (√I1 - √I2)² = (2√I0 - √I0)² = I0 (Imax / Imin) = 9 / 1	Wave Optics
Q7: In Young's double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are	(a) 1.56 mm (b) 7.8 mm (c) 9.75 mm (d) 15.6 mm	(b) 7.8 mm	Let y be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides. Now, for λ1, y = mλ1D/d For λ2, y = nλ2D/d mλ1 = nλ2 (m/n) = λ2/λ1 = 520/650 = 4/5 i.e. with respect to central maximum 4th bright fringe of λ1 coincides with 5th bright fringe of λ2 Now, y = (4 x 650 x 10^{-9} x 1.5) / (0.5 x 10^{-3}) m y = 7.8 x 10^{-3} m or y = 7.8 mm	Wave Optics
Q8: A single slit of width b is illuminated by coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)	(a) 6.0 cm (b) 1.5 cm (c) 4.5 cm (d) 3.0 cm	(b) 1.5 cm	For single slit diffraction, sinθ = nλ/b (n = 1,2,3----) sinθ ≈ θ ≈ tanθ b tanθ = nλ b tanθ1 = 2λ b(y1/D) = 2λ tanθ2 = 2λ b(y2/D) = 2λ (y2 - y1) b / D = 2λ (6 - 3) b / D = 2λ 3b / D = 2λ λD / b = 3/2 = 1.5 cm	Wave Optics
Q9: Unpolarized light of intensity I0 is incident on the surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?	(a) Transmitted light is partially polarized with intensity I0 / 2 (b) Transmitted light is completely polarized with intensity less than I0 / 2 (c) The reflected light is completely polarized with intensity less than I0 / 2 (d) The reflected light is partially polarized with intensity I0 / 2	(c) The reflected light is completely polarized with intensity less than I0 / 2	At Brewster's angle, i = tan⁻¹(μ), the reflected light is completely polarized, whereas refracted light is partially polarized. Thus, the reflected ray will have lesser intensity compared to the refracted ray.	Wave Optics
Q10: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is	(a) I0 / 8 (b) I0 (c) I0 / 2 (d) I0 / 4	(d) I0 / 4	The intensity of light after passing polaroid A is I1 = I0 / 2 Now this light will pass through the second polaroid B whose axis is inclined at an angle of 45° to the axis of polaroid A. So in accordance with Malus law, the intensity of light emerging from polaroid B is I2 = I1 cos² 45 = (I0 / 2)(1/√2)² = I0 / 4	Wave Optics
Q1: When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is	(a) 0.67 H (b) 1.67 H (c) 3 H (d) 6 H	(b) 1.67 H	I1=5 A, I2=2 A, ΔI=2-5=-3 A, Δt=0.1, S=50 V, As, ε=-L(ΔI / Δt), 50=-L(-3 / 0.1), 50=30 L, L=5 / 3=1.67 H	Electromagnetic Induction
Q2: Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A= 10 cm^2 and length =20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ0=4 π × 10^-7 T mA^-1)	(a) 2.4 π × 10^-4 H (b) 2.4 π × 10^-5 H (c) 4.8 π × 10^-4 H (d) 4.8 π × 10^-5 H	(a) 2.4 π × 10^-4 H	A=π r1^2=10 cm^2, l=20 cm, N1=300, N2=400, M=μ0 N1 N2 / l, M=(4 π × 10^-7 × 300 × 400 × 10 × 10^-4) / 0.20=2.4 π × 10^-4 H	Electromagnetic Induction
Q3: The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of	(a) 1 μF (b) 2 μF (c) 4 μF (d) 8 μF	(a) 1 μF	For maximum power, L ω=1 / C ω, C=1 / L ω^2, C=1 /[10 ×(2 π × 50)^2]=1 /(10 × 10^4 ×(π)^2)=10^-6 F=1 μF (Taking π^2=10)	Electromagnetic Induction
Q4: A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in	(a) 0.15 s (b) 0.3 s (c) 0.05 s (d) 0.1 s	(d) 0.1 s	During growth of charge in an inductance, I=I0(1-e^(-Rt / L)), or I0 / 2=I0(1-e^(-Rt / L)), e^(-Rt / L)=1 / 2=2^-1, or Rt / L=ln 2 ⇒ t=(L / R) ln 2, t=[(300 × 10^-3) / 2] ×(0.693), t=0.1 sec	Electromagnetic Induction
Q5: In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to	(a) 4 L (b) 2 L (c) L / 2 (d) L / 4	(c) L / 2	At resonance, ω=1 / √(L C) when ω is constant, (1 / L1 C1)=(1 / L2 C2)=(1 / LC)=1 / L2(2 C)=1 / 2 L2 C, L2=L / 2	Electromagnetic Induction
Q6: Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon	(a) the rates at which currents are changing in the two coils (b) relative position and orientation of the two coils (c) the materials of the wires of the coils (d) the currents in the two coils	(c) the materials of the wires of the coils	Mutual inductance between two coils depends on the materials of the wires of the coils	Electromagnetic Induction
Q7: The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is	(a) R / ω L (b) R / (R^2+ω^2 L^2)^(1 / 2) (c) ω L / R (d) R / (R^2-ω^2 L^2)^(1 / 2)	(b) R / (R^2+ω^2 L^2)^(1 / 2)	Power Factor, cos φ=1 / (1+tan^2 φ)^(1 / 2), cos φ=1 / (1+(ω L / R)^2)^(1 / 2)(because tan φ=ω L / R), cos φ=R / (R^2+ω^2 L^2)^(1 / 2)	Electromagnetic Induction
Q8: An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to	(a) 80 H (b) 0.08 H (c) 0.044 H (d) 0.065 H	(d) 0.065 H	Given I=10 A, V=80 V, R=V / I=80 / 10=8 Ω and ω=50 Hz, I=V / (8^2+X_L^2)^(1 / 2), 10=220 / (64+X_L^2)^(1 / 2), (64+X_L^2)^(1 / 2)=22, Squaring on both sides, we get 64+X_L^2=484, X_L^2=484-64=420, X_L=(420)^(1 / 2) ⇒ 2 π × ω L=(420)^(1 / 2), Series inductor on an arc lamp, L=(420)^(1 / 2) / (2 π × 50)=0.065 H	Electromagnetic Induction
Q9: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90 %, the output current would be	(a) 50 A (b) 25 A (c) 45 A (d) 20 A	(c) 45 A	Efficiency η=0.9=Ps / Pp, Vs Is=0.9 × Vp Ip, Is=(0.9 × 2300 × 5) / 230=45 A	Electromagnetic Induction
Q10: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be	(a) (3 / 2) nBA ω (b) nBA (c) 3 nBA ω (d) (1 / 2) nBA ω	(b) nBA	Emf induced in the coil is given by ε=BAn ω sin ω t, εmax=BA (The induced emf is maximum when sin ω t=1)	Electromagnetic Induction
Q11: A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth's magnetic field of 0.3 × 10^-4 Wb/m^2. The value of the induced emf in wire is	(a) 0.3 × 10^-3 V (b) 2.5 × 10^-3 V (c) 1.5 × 10^-3 V (d) 1.1 × 10^-3 V	(c) 1.5 × 10^-3 V	The motional emf is given as ε=\|v(l × B)\| =5 ×(0.3 × 10^-4)(10) sin 90°=1.5 × 10^-3 V	Electromagnetic Induction
Q12: A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil	(a) increases by a factor of 27 (b) decreases by a factor of 3 (c) increases by a factor of 3 (d) decreases by a factor of 9	(c) increases by a factor of 3		Electromagnetic Induction
Q13: The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is	(a) 637.5 J (b) 540 J (c) 437.5 J (d) 740 J	(c) 437.5 J	ε=25 V, I1=10 A, I2=25 A, t=1 s, ΔE=?, ε=L(ΔI / Δt), 25=L[(25-10) / 1], L=25 / 15=(5 / 3) H, ΔE=1 / 2 L(I2^2-I1^2)=(1 / 2) ×(5 / 3) ×(25^2-10^2)=(5 / 3) × 525=437.5 J	Electromagnetic Induction
Q14: There are two long co-axial solenoids of the same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self inductance of the inner coil is	(a) n2 / n1 (b) (n2 / n1)(r2^2 / r1^2) (c) (n2 / n1)(r1 / r2) (d) n1 / n2	(a) n2 / n1	The mutual inductance of the inner coil M=μ0 n1 n2 r1^2, Self inductance (L) of the inner coil is L=μ0 n1^2 r1^2, Using (1) and (2), M / L=n2 / n1	Electromagnetic Induction
Q15: A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is	(a) 2 mV (b) 6 mV (c) 1 mV (d) 12 mV	(d) 12 mV	Emf developed across the given edges, ε=Blv=0.1 × 0.02 × 6=12 × 10^-3 V=12 mV, As all the edges are parallel between the faces perpendicular to the x-axis, hence required potential difference is 12 mV	Electromagnetic Induction
Q1: If the distance between the earth and the sun were half its present value, the number of days in a year would have been	(a) 64.5 (b) 129 (c) 182.5 (d) 730	(b) 129 days	From Kepler's law, T² ∝ R³ Therefore, (T₂/T₁)² = (R₂/R₁)³ T₁ = 365 days, R₁ = R, R₂ = R/2 (T₂/365)² = (R/2/R)³ T₂² = (365)²/8 T₂² = 16,653 T₂ = 129 days	Gravitation
Q2: The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g' and 'R' (radius of the earth) is 10 m/s² and 6400 km respectively. The required energy for this work will be	(a) 6.4 × 10¹⁰ Joules (b) 6.4 × 10¹¹ Joules (c) 6.4 × 10⁸ Joules (d) 6.4 × 10⁰ Joules	(a) 6.4 × 10¹⁰ Joules	The energy required is given by = GMm/R = gR² × m/R (because g = GM/R²) = mgR = 1000 × 10 × 6400 × 10³ = 64 × 10⁹ J = 6.4 × 10¹⁰ J	Gravitation
Q3: An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) energy E₀. Its potential energy is	(a) -E₀ (b) 1.5E₀ (c) 2E₀ (d) E₀	(c) 2E₀	Total Energy, E₀ = -GMm/2r Potential Energy, U = -GMm/r = 2E₀	Gravitation
Q4: The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is	(a) R/2 (b) R/3 (c) 2R (d) 3R	(c) 2R	Acceleration due to gravity at a height "h" is given by g' = g(R/R + h)² Here, g is the acceleration due to gravity on the surface R is the radius of the earth As g' is given as g/9, we get g/9 = g(R/R + h)² 1/3 = R/R + h h = 2R	Gravitation
Q5: A simple pendulum has a time period T₁ when on the earth's surface, and T₂ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T₂/T₁ is	(a) 1 (b) 3 (c) 4 (d) 2	(d) 2	2π√(l/g) The time period of a simple pendulum = On the surface of earth g = GM/R² At a height R above the earth g = GM/(2R)² The time period on the surface of the earth T₁ = 2π√(lR²/GM) The time period on the surface of the earth T₂ = 2π√(l(2R)²/GM) T₂/T₁ = 2	Gravitation
Q6: A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (R = 6,400 km) will approximately be	(a) 1/2 hr (b) 1 hr (c) 2 hr (d) 4 hr	(c) 2 hr	According to Kepler's law, T² ∝ R³ Therefore, (T₂/T₁)² = (R₂/R₁)³ For a spy satellite time period is given by T₁ R₁ = 6400 km For a geostationary satellite, T₂ = 24 hours R₂ = 36,000 km (24/T₁)² = (36000/6400)³ (24/T₁)² = 178 (24/T₁) = 13.34 T₁ = 24/13.34 ≈ 2 hr	Gravitation
Q7: Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is	(a) -6 Gm/r (b) -9 Gm/r (c) Zero (d) -4 Gm/r	(b) -9 Gm/r	P is the point where the field is zero, and a unit mass is placed at P. Applying Newton's law of gravitation, (Gm × 1)/x² = (G 4m × 1)/(r - x)² x²/(r - x)² = 1/4 x/(r - x) = 1/2 2x = r - x x = r/3 Potential at the point P, V = -Gm/x - G(4m)/(r - x)	Gravitation
Q8: A Binary star system consists of two stars A and B which have time periods Tₐ and Tᵦ, radii Rₐ and Rᵦ and masses Mₐ and Mᵦ. Then	(a) Tₐ > Tᵦ then Rₐ > Rᵦ (b) Tₐ > Tᵦ then Mₐ > Mᵦ (c) (Tₐ/Tᵦ)² = (Rₐ/Rᵦ)² (d) Tₐ = Tᵦ	(d) Tₐ = Tᵦ	Angular velocity of binary stars are the equal ωₐ = ωᵦ 2π/Tₐ = 2π/Tᵦ ∴ Tₐ = Tᵦ	Gravitation
Q9: Two particles of equal mass "m" go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is	(a) √(Gm/R) (b) √(Gm/4R) (c) √(Gm/3R) (d) √(Gm/2R)	(b) √(Gm/4R)	Gm²/4R² = mv²/R V = √(Gm/4R)	Gravitation
Q10: A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass "m" is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is	(a) 1/2 mV² (b) mV² (c) 3/2 mV² (d) 2 mV²	(b) mV²	Kinetic energy at the time of ejection = 1/2 mvₑ² Vₑ is the escape velocity = √2 × orbital velocity Vₑ is the escape velocity = √2 v Kinetic energy at the time of ejection = 1/2 m(√2 v)² = mv²	Gravitation
Q11: Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (F = Gm₁m₂/r²)		V = √(Gm/a) and T = 2π √(a³/3 Gm)	ABC is an equilateral triangle of side a Each particle moves in a circle of radius r AD² = a² - a²/4, r = 2/3 AD r = 2/3 × √(a² - a²/4) r = a √3---(1) To find v Let v = Initial velocity given to each particle. For circular motion, centripetal force should be provided. It is provided by the gravitational force between two masses. Let F denote this force. F = Gm²/a² Resultant force = √(F² + F² + 2 F² cos 60°) Resultant force = √3 F Resultant force = Centripetal force √3 F = mv²/r V² = √3 F r/m = √3/m × (Gm²/a²) (a/√3) = Gm/a V = √(Gm/a) To find time period of circular motion(T) T = 2πr/V = 2π(a/√3) × √(a/ Gm) = 2π √(a³/3 Gm) therefore, T = 2π √(a³/3 Gm)	Gravitation
Q12: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?	(a) 5 GmM/6 R (b) 2 GmM/3 R (c) GmM/2 R (d) GmM/3 R	(a) 5 GmM/6 R	Energy of the satellite on the surface of the Earth E₁ = -GMm/R Energy at a distance 2R is given by E₂ = -GMm/3R + 1/2 mv₀² E₂ = -GMm/3R + 1/2 m[GM/3R] E₂ = -GmM/6R E₂ - E₁ = (-GmM/6R) - (-GMm/R) = 5GmM/6R	Gravitation
Q13: Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours, respectively. The radius of the orbit of S1 is 10⁴ km. When S2 is closest to S1, find the angular speed of S2 as actually observed by an astronaut in S1		Angular speed = 7π/8	Angular velocity ω = 2π/T Here T = 1 hour for S₁, ω₁ = 2π rad/hr Similarly, for S₂, ω₂ = 2π Given that they are rotating in the same sense. So, relative angular velocity = 2π - 2π/8 = 7π/8	Gravitation
Q14: Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence	(a) There will be no change in weight anywhere on the earth (b) Weight of the object, everywhere on the earth, will increase (c) Except at poles, weight of the object on the earth will decrease (d) Weight of the object, everywhere on the earth, will decrease.	(c) Except at poles, weight of the object on the earth will decrease	The effect of rotation of earth on acceleration due to gravity is given by g' = g - ω²R cos 2Φ Where Φ is latitude. There will be no change in gravity at poles as Φ = 90°, at all points as ω increases g' will decrease.	Gravitation
Q15: A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nᵗʰ power of R. If the period of rotation of the particle is T, then	(a) T ∝ R³/₂ (b) T ∝ R(m 2)+1 (c) T ∝ R(n+1)/2 (d) T ∝ Rⁿ/2	(c) T ∝ R(n+1)/2	According to the question, central force is given by Fᶜ ∝ 1/Rⁿ Fᶜ = k(1/Rⁿ) mω²R = k(1/Rⁿ) m(2π)²/T² = k(1/Rⁿ+¹) Or T² ∝ Rⁿ+¹ T ∝ R(n+1)/2	Gravitation
Q1: A solid sphere of radius $R$ acquires a terminal velocity $v_{1}$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into 27 identical spheres. If each of these acquires a terminal velocity $v_{2}$, when falling through the same fluid, the ratio ( $\left.v_{1} / v_{2} ight)$ equals	(a) 9 (b) $1 / 27$ (c) $1 / 9$ (d) 27	(a) 9	$27 \mathrm{x}(4 / 3) \pi \mathrm{r}^{3}=(4 / 3) \pi \mathrm{R}^{3}$ $\operatorname{Or} \mathrm{r}=\mathrm{R} / 3$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{3}$ Therefore, $\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight)=\left(\mathrm{R}^{2} / \mathrm{r}^{2} ight)$ $\mathrm{v}_{1} \mathrm{v}_{2}=[\mathrm{R} /(\mathrm{R} / 3)]^{2}=9$ $\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight)=9$	Fluid Mechanics
Q2: Spherical balls of radius $R$ are falling in a viscous fluid of viscosity with a velocity $v$. The retarding viscous force acting on the spherical ball is	(a) directly proportional to $\mathrm{R}$ but inversely proportional to $\mathrm{v}$ (b) directly proportional to both radius $\mathrm{R}$ and velocity $\mathrm{v}$ (c) inversely proportional to both radius $\mathrm{R}$ and velocity $\mathrm{v}$ (d) inversely proportional to $\mathrm{R}$ but directly proportional to velocity $\mathrm{v}$	(b) directly proportional to both radius $R$ and velocity $v$	Retarding viscous force $=6 \pi \eta R v$ obviously option (b) holds goods	Fluid Mechanics
Q3: A long cylindrical vessel is half-filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is $5 \mathrm{~cm}$ and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in $\mathrm{cm}$, will be	(a) 0.4 (b) 2.0 (c) 0.1 (d) 1.2	(b) 2.0	The linear speed of the liquid at the sides is $\mathrm{r} \omega$. So, the difference in height is given as follows $2 \mathrm{gh}=\omega^{2} \mathrm{r}^{2}$ $\mathrm{h}=\omega^{2} \mathrm{r}^{2} / 2 \mathrm{~g}$ here $\omega=2 \pi \mathrm{f}$ Therefore, $\mathrm{h}=\left[(2 imes 2 \pi)^{2}\left(5 imes 10^{-2} ight)^{2} ight] /(2 imes 10)=2 \mathrm{~cm}$	Fluid Mechanics
Q4: Water is flowing continuously from a tap having an internal diameter $8 imes 10^{-3} \mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 imes 10^{-1} \mathrm{~m}$ below the tap is close to	(a) $5.0 imes 10^{-3} \mathrm{~m}$ (b) $7.5 imes 10^{-3} \mathrm{~m}$ (c) $9.6 imes 10^{-3} \mathrm{~m}$ (d) $3.6 imes 10^{-3} \mathrm{~m}$	(d) $3.6 imes 10^{-3} \mathrm{~m}$	Here, $\mathrm{d}_{1}=8 imes 10^{-3} \mathrm{~m}$ $\mathrm{v}_{1}=0.4 \mathrm{~m} \mathrm{~s}^{-1}, \mathrm{~h}=0.2 \mathrm{~m}$ According to equation of motion, $$ \begin{aligned} & \quad v_{2}=\sqrt{v_{1}^{2}+2 g h}=\sqrt{(0.4)^{2}+2+10 imes 0.2} \n& =2 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$ According to equation of continuity $\mathrm{a}_{1} \mathrm{v}_{1}=\mathrm{a}_{2} \mathrm{v}_{2}$ According to equation of continuity $\mathrm{a}_{1} \mathrm{v}_{1}=\mathrm{a}_{2} \mathrm{v}_{2}$ $\left(\pi \mathrm{D}_{1}{ }^{2} / 4 ight) \mathrm{v}_{1}=\left(\pi \mathrm{D}_{2}{ }^{2} / 4 ight) \mathrm{v}_{2}$ $\mathrm{D}_{2}{ }^{2}=\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight) \mathrm{D}_{1}{ }^{2}$ $\mathrm{D}_{2}=\left[\sqrt{ }\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight) ight] \mathrm{D}_{1}$ $=[\sqrt{ }(0.4 / 2)] imes 8 imes 10^{-3} \mathrm{~m}$ $\mathrm{D}_{2}=3.6 imes 10^{-3} \mathrm{~m}$	Fluid Mechanics
Q5: A $20 \mathrm{~cm}$ long capillary tube is dipped in water. The water rises up to $8 \mathrm{~cm}$. If the entire arrangement is put in a freely falling elevator the length of the water column in the capillary tube will be	(a) $4 \mathrm{~cm}$ (b) $20 \mathrm{~cm}$ (c) $8 \mathrm{~cm}$ (d) $10 \mathrm{~cm}$	(b) $20 \mathrm{~cm}$	In a freely falling elevator, $\mathrm{g}=0$ Water will rise to the full length i.e., $20 \mathrm{~cm}$ to tube	Fluid Mechanics
Q6: A spherical solid ball of volume $V$ is made of a material of density $ ho_{1}$. It is falling through a liquid of density $ ho_{2}\left( ho_{2}<1 ight)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{ ext {viscous }}=-\mathbf{k v}^{2}(k>0)$. The terminal speed of the ball is	(a) $\operatorname{Vg}\left( ho_{1}- ho_{2} ight)$ (b) \[ \sqrt{ \frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}} \] (c) $\operatorname{Vg} ho_{1} / \mathrm{k}$ (d) $\sqrt{\frac{V g ho_{1}}{k}}$	(b) \[ \sqrt{ \frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}} \]	The forces acting on the solid ball when it is falling through a liquid is "mg" downwards, thrust by Archimedes principle upwards and the force due to the force of friction also acting upwards. The viscous force rapidly increases with velocity, attaining a maximum when the ball reaches the terminal velocity. Then the acceleration is zero $\mathrm{mg}-\mathrm{V} ho_{2} \mathrm{~g}-\mathrm{kv}_{\mathrm{t}}^{2}=$ ma where $\mathrm{V}$ is volume, $\mathrm{v}_{\mathrm{t}}$ is the terminal velocity When the ball is moving with terminal velocity, $\mathrm{a}=0$ Therefore $\mathrm{V} ho_{1} \mathrm{~g}-\mathrm{V} ho_{2} \mathrm{~g}-\mathrm{kv}_{\mathrm{t}}^{2}=0$ $\mathrm{v}_{\mathrm{t}}=\sqrt{\frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}}$	Fluid Mechanics
Q7: Water flows into a large tank with a flat bottom at the rate of $10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1}$. Water is also leaking out of a hole of area $1 \mathrm{~cm}^{2}$ at its button. If the height of the water in the tank remains steady, then this height is	(a) $5 \mathrm{~cm}$ (b) $7 \mathrm{~cm}$ (c) $4 \mathrm{~cm}$ (d) $9 \mathrm{~cm}$	(a) $5 \mathrm{~cm}$	Since the height of the water column is constant Water inflow rate $\left(\mathrm{Q}_{ ext {in }} ight)=$ Water outflow rate $\left(\mathrm{Q}_{ ext {out }} ight)$ $\mathrm{Q}_{ ext {in }}=10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1}$ $\mathrm{Q}_{ ext {out }}=10^{-4} \mathrm{x} \sqrt{ }(2 \mathrm{gh})$ $10^{-4}=10^{-4} \mathrm{x} \sqrt{ } 20 \mathrm{xh}$ $\mathrm{h}=(1 / 20) \mathrm{m}=5 \mathrm{~cm}$	Fluid Mechanics
Q8: A submarine experiences a pressure of $5.05 imes 10^{6} \mathrm{~Pa}$ at depth of $d_{1}$ in a sea. When it goes further to a depth of $d_{2}$, it experiences a pressure of $8.08 imes 10^{6} \mathrm{~Pa}$. Then $d_{1}-d_{2}$ is approximately (density of water $=10^{3}$ $\mathrm{ms}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )	(a) $300 \mathrm{~m}$ (b) $400 \mathrm{~m}$ (c) $600 \mathrm{~m}$ (d) $500 \mathrm{~m}$	(a) $300 \mathrm{~m}$	$\mathrm{P}_{1}=\mathrm{P}_{0}+ ho \mathrm{gd}_{1}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+ ho \mathrm{gd}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}= ho g \Delta \mathrm{d}$ $\left(8.08 imes 10^{6}-5.05 imes 10^{6} ight)=10^{3} imes 10 imes \Delta \mathrm{d}$ $3.03 imes 10^{6}=10^{3} imes 10 imes \Delta \mathrm{d}$ $\Delta \mathrm{d}=303 \mathrm{~m} pprox 300 \mathrm{~m}$	Fluid Mechanics
Q9: Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is $5 \mathrm{~cm}$, the Reynolds number for the flow is of the order (density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$, coefficient of viscosity of water $=1 \mathrm{mPa} \mathrm{s}$ )	(a) $10^{3}$ (b) $10^{4}$ (c) $10^{2}$ (d) $10^{6}$	(b) $10^{4}$	Reynolds number $= ho v d / \eta$ Volume flow rate $=\mathrm{vx} \pi \mathrm{r}^{2}$ $\mathrm{v}=\left(100 imes 10^{-3} / 60 ight) imes\left(1 / \pi imes 25 imes 10^{-4} ight)$ $\mathrm{v}=(2 / 3 \pi) \mathrm{m} / \mathrm{s}$ Reynolds number $=\left\{\left(10^{3} imes 2 imes 10 imes 10^{-2} ight) /\left(10^{-3} imes 3 \pi\right)\right\}$ $\simeq 2 imes 10^{4}$ Order of $10^{4}$	Fluid Mechanics
Q10: The top of a water tank is open to the air and its water level is maintained. It is giving out $0.74 \mathrm{~m}^{3}$ water per minute through a circular opening of $2 \mathrm{~cm}$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to	(a) $6.0 \mathrm{~m}$ (b) $4.8 \mathrm{~m}$ (c) $9.6 \mathrm{~m}$ (d) $2.9 \mathrm{~m}$	(b) $4.8 \mathrm{~m}$	Here, volumetric flow rate $=(0.74 / 60)=\pi r^{2} v=\left(\pi imes 4 imes 10^{-4} ight) imes \sqrt{ } 2$ gh $\Rightarrow \sqrt{ } 2 \mathrm{gh}=[(74 imes 100) / 240 \pi)]$ $\Rightarrow \sqrt{ } 2 \mathrm{gh}=740 / 24 \pi$ $2 \mathrm{gh}=(740 / 24 \pi)^{2}$ $\mathrm{h}=[(740 imes 740) / 24 imes 24 imes 10)]\left(\right.$ since $\left.\pi^{2}=10\right)$ $\mathrm{h} pprox 4.8 \mathrm{~m}$	Fluid Mechanics
Q1: During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts	(a) in the backward direction on the front wheel and in the forward direction on the rear wheel (b) in the forward direction on the front wheel and in the backward direction on the rear wheel (c) in the backward direction on both, the front and the rear wheels (d) in the forward direction on both, the front and the rear wheels.	Answer: (a) Due to peddling, the point of contact of the rear wheel has a tendency to move backwards. So frictional force opposes the backwards tendency i.e., the frictional force acts in the forward direction. But the back wheel accelerates the front wheel in the forward direction. To oppose this frictional force acts in the backward direction on the front wheel.		Friction
Q2: A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5 , the magnitude of the frictional force acting on the block is	(a) 2.5 N (b) 0.98 N (c) 4.9 N (d) 0.49 N	Answer: (a) 2.5 N	Solution: Consider the forces, acting on the block in the vertical direction Force of friction f=μR f=0.5×5 f=2.5 N	Friction
Q3: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2 . The weight of the block is-	(a) 20 N (b) 50 N (c) 100 N (d) 2 N	Answer: (d) 2N	Solution: Frictional force balances the weight of the body Frictional force f=μN=mg f=0.2X 10=2 N Therefore, weight of the block mg=2 N	Friction
Q4: A marble block of mass 2 kg lying on ice when given a velocity of 6 m / s is stopped by friction in 10 s. Then the coefficient of friction is (consider g=10 m / s)	(a) 0.02 (b) 0.03 (c) 0.06 (d) 0.01	Answer: (c) 0.06	Solution: u=6 m / s v=0 t=10 s a=-f / m=-μmg / m=-μg=-10 μ Substituting values in v=u+at 0=6-10 μ× 10 Therefore, μ=0.06	Friction
Q5: A block P of mass m is placed on a horizontal frictionless plane. The second block of the same mass m is placed on it and is connected to a spring of spring constant k. The two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of the frictional force between the two block.	(a)kA/2 (b) kA (c) μs mg (d) zero	Answer: (a) fm=k A / 2	Solution: Block Q oscillates but does not slip on P. It means that acceleration is the same for Q and P both. There is a force of friction between the two blocks while the horizontal plane is frictionless. The spring is connected to the upper block. The (P-Q) system oscillates with angular frequency ω. The spring is stretched by A. Therefore, ω=√k / 2 m Therefore, Maximum acceleration in SHM=ω^2 A a_m=kA / 2 m Now consider the lower block. Let the maximum force of friction =f_m f_m=ma_m or f_m=mx(kA / 2 m) f_m=kA / 2	Friction
Q6: Statement 1: A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in mechanical energy in the second situation is smaller than that in the first situation.	Statement-2: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. (a) Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1 (b) Statement-1 and 2 are true and statement-2 is not a correct explanation for statement-1 (c) Statement-1 is true, statement-2 is false (d) Statement-1 is false, statement-2 is true.	Answer: (c) Statement-1 is true, statement-2 is false		Friction
Q7: A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8 . If the frictional force on the block is 10 N, the mass of the block (in kg) is : (taken g=10 m / s² )	(a) 2.0 (b) 4.0 (c) 1.6 (d) 2.5	Answer: (a) 2	Solution: μ=0.8 f=10 N mg sin 30=10 mx 10×(1 / 2)=10 m=2 kg	Friction
Q8: A smooth block is released at rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on a rough incline than on a smooth incline. The coefficient of friction is-	(a) μ k=1-1 / n² (b) μ k=√(1-1 / n²) (c) μ s=1-1 / n² (d) μ s=√(1-1 / n²)	Answer: (a) μ k=1-1 / n²	Solution: On a smooth plane d=1 / 2(g sin θ) t_1 t_1=√(2 d / g sin θ) On a rough surface d=1 / 2(g sin θ-μ g cos θ) t_2 t_2=√(2 d / g sin θ-μ g cos θ) From the question, we know t_2=nt_1 √(2 d / g sin θ-μ g cos θ) n√(2 d / g sin θ) n=1 / √(1-μ_k) (since cos 45°=sin 45° 1 / √2) n²=1 / 1-μ_k 1-μ_k=1 / n² μ_k=1-1 / n²	Friction
Q9: The upper half of an inclined plane with inclination Φ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by	(a) 2 sin Φ (b) 2 cos Φ (c) 2 tan Φ (d) tan Φ	Answer: (c) 2 tan Φ	Solution: Suppose the length of the plane is L. When the block slides down the plane, the increase in the K.E will be equal to the decrease in the P.E Work done, W= Change in K.E(ΔK)=(1 / 2) mu²-(1 / 2) mv²=0 Work done by friction (W_f)+ Work done by gravity (W_g)=0 -μ g cos Φ(L / 2)+mgLsin Φ=0 (μ / 2) cos Φ=sin Φ μ=2 tan Φ	Friction
Q10: Consider a car moving on a straight road with a speed of 100 m / s. The distance at which a car can be stopped is : (μ_k=0.5)	(a) 800 m (b) 1000 m (c) 100 m (d) 400 m	Answer: (b) 1000	Solution: When the car is stopped by friction then its retarding force is ma=μ R ma=μ mg a=μ g Consider the equation v²=u²-2 as (the car is retarding so a is negative) 0=u²-2 as 2as =u² s=u² / 2 a s=u² / 2 μ g s=(100)² / 2× 0.5× 10 s=1000 m	Friction
Q11: The minimum force required to start pushing a body up a rough (frictional coefficient μ ) inclined plane F₁ while the minimum force needed to prevent it from sliding down is F₂. If the inclined plane makes an angle θ from the horizontal such that tan θ=2 μ then the ratio F₁ / F₂ is	(a) 4 (b) 1 (c) 2 (d) 3	Answer: (d) 3	Solution: We have to work against the sliding force due to gravity and frictional force to push upwards. Therefore, the force F₁=mg sin θ+μ mg cos θ. To stop the body from sliding work is done against sliding force but frictional force stops the body from sliding. Therefore, the force F₂=mg sin θ-μ mg cos θ. F₁ / F₂=(mg sin θ+μ mg cos θ) /(mg sin θ-μ mg cos θ) F₁ / F₂=(tan θ+μ) /(tan θ-μ) (since tan θ=2 μ ) F₁ / F₂=(2 μ+μ) /(2 μ-μ) F₁ / F₂=3	Friction
Q12: A body of mass m=10⁻² kg is moving in a medium and experiences a frictional force F=-kv². Its initial speed is v₀=10 ms⁻¹. If, after 10 s, its energy is 1 / 8 mv₀², the value of k will be	(a) 10⁻³ kg m⁻¹ (b) 10⁻³ kg s⁻¹ (c) 10⁻⁴ kg m⁻¹ (d) 10⁻¹ kg m⁻¹ s⁻¹	Answer: (c) 10⁻⁴ kg m⁻¹	Solution: Mass m=10² Kg Initial velocity v₀=10 ms⁻¹ Time t=10 s Frictional Force F=-kv² 1 / 2 mv_f²=1 / 8 mv₀² v_f=v₀ / 2 v_f=10 / 2=5 m / s Now, the force is (10⁻²) dv / dt=-kv² ∫_{10}^{5} (dv / v²)=-100 k ∫_{0}^{10} (dt) 1 / 5=1 / 10=100 kx 10 k=10⁻⁴	Friction
Q13: A block of mass m is placed on a surface with a vertical cross-section given by y=x³ / 6. If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is	(a) 1 / 3 m (b) 1 / 2 m (c) 1 / 6 m (d) 2 / 3 m	Answer: (c) 1 / 6	Solution: For equilibrium under limiting friction mg sin θ=μ mg cos θ tan θ=μ The equation of the surface is given by y=x³ / 6 Slope dy / dx=x² / 2 From (1) and (2) we get μ=x² / 2=0.5 x=1 So, y=1 / 6	Friction
Q2: The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is	(a) 2 / 5 (b) 3 / 2 (c) 3 / 5 (d) 2 / 3	(a) 2 / 5	For an ideal gas in an isobaric process, Heat supplied, Q=nCp ΔT Work done, W=P ΔV=nR ΔT Required Ratio =W / Q Required Ratio =(nR ΔT)/(nCp ΔT) R / Cp=R/(γR / γ-1) [since γ=5 / 3, for a monoatomic gas ] =2 / 5	Thermal Expansion
Q3: The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is (R=8.3 J mol−1 K−1)	(a) monoatomic (b) diatomic (c) triatomic (d) a mixture of monoatomic and diatomic	(b) diatomic	Increase in temperature, ΔT=T2−T1=7 K (Since it is change in temperature so its value will be same in Kelvin and degree celsius) Work done on the system, W=146 KJ=146 × 1000 J Number of moles of gas, n=1000 Work done in an adiabatic process, W=nR(T2−T1)/γ-1 put the values, 146000=1000 × 8.3 × 7 /(γ-1) γ-1=0.39 γ≈1.4 Hence the gas is diatomic	Thermal Expansion
Q4: Two moles of an ideal monatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.	(a) (i) 198 K (ii) 2.7 kJ (b) (i) 195 K (ii) 2.7 kJ (c) (i) 189 K (ii) 2.7 kJ (d) (i) 195 K (ii) 2.7 kJ	(c) (i) 189 K (ii) 2.7 kJ	For an adiabatic process, PV^γ=constant (nRT/V)V^γ=constant or TV^(γ-1)=constant T1 V1^(γ-1)=T2 V2^(γ-1) T2=T1[V1 / V2]^(γ-1) Here, T1=27°C=300 K, V1=V, V2=2 V, γ=5 / 3 T2=300(V / 2 V)(5 / 3-1)=300(1 / 2)^(2 / 3)=189 K Change in internal energy, ΔU=nCv ΔT ΔU=n[(f / 2)R](T2−T1)=2 ×(3 / 2) ×(25 / 3)(189-300) ΔU=-2.7 kJ	Thermal Expansion
Q6: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 1000°C is (For steel Young's modulus is 2 × 10^11 Nm−2 and coefficient of thermal expansion is 1.1 × 10−5 K−1)	(a) 2.2 × 10^7 Pa (b) 2.2 × 10^6 Pa (c) 2.2 × 10^8 Pa (d) 2.2 × 10^9 Pa	(c) 2.2 × 10^8 Pa	Given, ΔT=100°C, Y=2 × 10^11 N m−2 α=1.1 × 10−5 K−1 Young's modulus, Y= stress/strain =(F / A) /ΔL / L Therefore, Y=P /ΔL / L Y=P / α ΔT [ since ΔL=L α ΔT] Thermal stress in a rod is the pressure due to the thermal strain. Required pressure P=Y α ΔT=2 × 10^11 × 1.1 × 10−5 × 100=2.2 × 10^8 Pa	Thermal Expansion
Q7: Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?	(a) 6 F (b) 9 F (c) F (d) 4 F	(d) 9 F	For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire 11 is A and that of 12 is 3 A. V=V1=V2 V=A × 11=3 A × 12 12=11 / 3 Y=(F / A) /Δ1 / l ⇒F1=YA(Δ11 / 11) F2=Y 3 A(Δl2 / l2) Given Δl1=Δl2=Δx (for the same extension) F2=Y 3 A(Δx / 11 / 3)=9(YA Δx / 11)=9 F1 or 9 F	Thermal Expansion
Q8: A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to	(a) 9 F / (π r^2 YT) (b) F / (3 π r^2 YT) (c) 6 F / (π r^2 YT) (d) 3 F / (π r^2 YT)	(d) 3 F / (π r^2 YT)	Given, Length =L Longitudinal force =F Radius =r Young's modulus =Y Temperature =T Since length remains same (Stress)Compressive =( Stress ) Thermal F / A=Y α T ⇒F / π r^2=Y α T (where α= coeficient of linear expansion) α=F / π r 2 YT Coefficient of volume expansion, γ=3 α γ=3 F / π r^2 YT	Thermal Expansion
Q9: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP / CV for the gas is	(a) 4 / 3 (b) 2 (c) 5 / 3 (d) 3 / 2	(d) 3 / 2	In an adiabatic process, T^γ= (constant) P^γ-1 T^γ / γ-1= (constant)P Given T^3= (constant) P γ / γ-1=3 ⇒ 3 γ-3=γ ⇒ γ=3 / 2	Thermal Expansion
Q10: An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by	(a) 3 / P α K (b) P / 3 α K (c) 3 α / PK (d) 3 PK α	(b) P / 3 α K	The bulk modulus of the gas is given by K=-P /ΔV / V 0 (Here negative sign indicates the decrease in volume with pressure) ΔV / V0=P / K [in magnitude] Also, V=V0(1+γ ΔT) or ΔV / V0=γ ΔT Comparing eq. (1) and (2), we get P / K=γ ΔT ΔT=P / 3 α K (since γ=3 α )	Thermal Expansion
Q11: Steel wire of length ' L ' at 40°C is suspended from the ceiling and then a mass ' m ' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length ' L '. The coefficient of linear thermal expansion of the steel is 10−5 /°C, Young's modulus of steel is 10^11 N / m^2 and radius of the wire is 1 mm. Assume that L > diameter of the wire. Then the value of ' m ' in kg is nearly		3	E=(F / A) /ΔL / L =(F × L)/(A × ΔL) ⇒(ΔL / L)=F / AE Since, (ΔL / L)=α Δt α Δt=F / AE α Δt=mg / AE m=[(α Δt) / g] AE m=3.14 ≈ 3	Thermal Expansion
Q12: A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?	(a) 25 J (b) 35 J (c) 30 J (d) 40 J	(b) 35 J	The value of Cp for a diatomic gas is Cp=(7 / 2) R-(1) It is given that work done during expansion at constant pressure is dW=10 J=nR ΔT Since the gas undergoes an isobaric process ⇒ ΔQ=n(7 / 2) R ΔT=7 / 2(nR ΔT)=(7 / 2)(10)=35 J	Thermal Expansion

Q4: In an interference experiment the ratio of amplitudes of coherent waves is (a1 / a2) = (1 / 3). The ratio of maximum and minimum intensities of fringes will be

(a) 4 (b) 9 (c) 2 (d) 18

(a) 4

(a1 / a2) = (1 / 3) Imax = a1 + a2 Imin = a1 - a2 (Imax / Imin) = [(a1 + a2)² / (a1 - a2)²] = [(1 + 1/3)² / (1 - 1/3)²] = (4/2)² = 4

Wave Optics

Q5: Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.

(a) 610 x 10^{-9} radian (b) 152.5 x 10^{-9} radian (c) 457.5 x 10^{-9} radian (d) 305 x 10^{-9} radian

(d) 305 x 10^{-9} radian

The limit of resolution, θΔ = 1.22λ/a = (1.22 x 500 x 10^{-9}) / (200 x 10^{-2}) = 3.05 x 10^{-7} radian = 305 x 10^{-9} radian

Wave Optics

Q6: In Young's double-slit experiment, the ratio of the slit's width is 4:1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be

(a) (√3 + 1)² : 16 (b) 9 : 1 (c) 25 : 9 (d) 4 : 1

(b) 9 : 1

I1 = 4I0 I2 = I0 Imax = (√I1 + √I2)² = (2√I0 + √I0)² = 9I0 Imin = (√I1 - √I2)² = (2√I0 - √I0)² = I0 (Imax / Imin) = 9 / 1

Wave Optics

Q7: In Young's double-slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide are

(a) 1.56 mm (b) 7.8 mm (c) 9.75 mm (d) 15.6 mm

(b) 7.8 mm

Let y be the distance from the central maximum to the point where the bright fringes due to both the wavelengths coincides. Now, for λ1, y = mλ1D/d For λ2, y = nλ2D/d mλ1 = nλ2 (m/n) = λ2/λ1 = 520/650 = 4/5 i.e. with respect to central maximum 4th bright fringe of λ1 coincides with 5th bright fringe of λ2 Now, y = (4 x 650 x 10^{-9} x 1.5) / (0.5 x 10^{-3}) m y = 7.8 x 10^{-3} m or y = 7.8 mm

Wave Optics

Q8: A single slit of width b is illuminated by coherent monochromatic light of wavelength λ. If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm, respectively from the central maximum, what is the width of the central maximum? (i.e. the distance between the first minimum on either side of the central maximum)

(a) 6.0 cm (b) 1.5 cm (c) 4.5 cm (d) 3.0 cm

(b) 1.5 cm

For single slit diffraction, sinθ = nλ/b (n = 1,2,3----) sinθ ≈ θ ≈ tanθ b tanθ = nλ b tanθ1 = 2λ b(y1/D) = 2λ tanθ2 = 2λ b(y2/D) = 2λ (y2 - y1) b / D = 2λ (6 - 3) b / D = 2λ 3b / D = 2λ λD / b = 3/2 = 1.5 cm

Wave Optics

Q9: Unpolarized light of intensity I0 is incident on the surface of a block of glass at Brewster's angle. In that case, which one of the following statements is true?

(a) Transmitted light is partially polarized with intensity I0 / 2 (b) Transmitted light is completely polarized with intensity less than I0 / 2 (c) The reflected light is completely polarized with intensity less than I0 / 2 (d) The reflected light is partially polarized with intensity I0 / 2

(c) The reflected light is completely polarized with intensity less than I0 / 2

At Brewster's angle, i = tan⁻¹(μ), the reflected light is completely polarized, whereas refracted light is partially polarized. Thus, the reflected ray will have lesser intensity compared to the refracted ray.

Wave Optics

Q10: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is

(a) I0 / 8 (b) I0 (c) I0 / 2 (d) I0 / 4

(d) I0 / 4

The intensity of light after passing polaroid A is I1 = I0 / 2 Now this light will pass through the second polaroid B whose axis is inclined at an angle of 45° to the axis of polaroid A. So in accordance with Malus law, the intensity of light emerging from polaroid B is I2 = I1 cos² 45 = (I0 / 2)(1/√2)² = I0 / 4

Wave Optics

Q1: When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self-inductance of the coil is

(a) 0.67 H (b) 1.67 H (c) 3 H (d) 6 H

(b) 1.67 H

I1=5 A, I2=2 A, ΔI=2-5=-3 A, Δt=0.1, S=50 V, As, ε=-L(ΔI / Δt), 50=-L(-3 / 0.1), 50=30 L, L=5 / 3=1.67 H

Electromagnetic Induction

Q2: Two coaxial solenoids are made by winding thin insulated wire over a pipe of cross-sectional area A= 10 cm^2 and length =20 cm. If one of the solenoids has 300 turns and the other 400 turns, their mutual inductance is (μ0=4 π × 10^-7 T mA^-1)

(a) 2.4 π × 10^-4 H (b) 2.4 π × 10^-5 H (c) 4.8 π × 10^-4 H (d) 4.8 π × 10^-5 H

(a) 2.4 π × 10^-4 H

A=π r1^2=10 cm^2, l=20 cm, N1=300, N2=400, M=μ0 N1 N2 / l, M=(4 π × 10^-7 × 300 × 400 × 10 × 10^-4) / 0.20=2.4 π × 10^-4 H

Electromagnetic Induction

Q3: The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of

(a) 1 μF (b) 2 μF (c) 4 μF (d) 8 μF

(a) 1 μF

For maximum power, L ω=1 / C ω, C=1 / L ω^2, C=1 /[10 ×(2 π × 50)^2]=1 /(10 × 10^4 ×(π)^2)=10^-6 F=1 μF (Taking π^2=10)

Electromagnetic Induction

Q4: A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady state value in

(a) 0.15 s (b) 0.3 s (c) 0.05 s (d) 0.1 s

(d) 0.1 s

During growth of charge in an inductance, I=I0(1-e^(-Rt / L)), or I0 / 2=I0(1-e^(-Rt / L)), e^(-Rt / L)=1 / 2=2^-1, or Rt / L=ln 2 ⇒ t=(L / R) ln 2, t=[(300 × 10^-3) / 2] ×(0.693), t=0.1 sec

Electromagnetic Induction

Q5: In a LCR circuit capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to

(a) 4 L (b) 2 L (c) L / 2 (d) L / 4

(c) L / 2

At resonance, ω=1 / √(L C) when ω is constant, (1 / L1 C1)=(1 / L2 C2)=(1 / LC)=1 / L2(2 C)=1 / 2 L2 C, L2=L / 2

Electromagnetic Induction

Q6: Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon

(a) the rates at which currents are changing in the two coils (b) relative position and orientation of the two coils (c) the materials of the wires of the coils (d) the currents in the two coils

(c) the materials of the wires of the coils

Mutual inductance between two coils depends on the materials of the wires of the coils

Electromagnetic Induction

Q7: The power factor of an AC circuit having resistance (R) and inductance (L) connected in series and an angular velocity ω is

(a) R / ω L (b) R / (R^2+ω^2 L^2)^(1 / 2) (c) ω L / R (d) R / (R^2-ω^2 L^2)^(1 / 2)

(b) R / (R^2+ω^2 L^2)^(1 / 2)

Power Factor, cos φ=1 / (1+tan^2 φ)^(1 / 2), cos φ=1 / (1+(ω L / R)^2)^(1 / 2)(because tan φ=ω L / R), cos φ=R / (R^2+ω^2 L^2)^(1 / 2)

Electromagnetic Induction

Q8: An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to

(a) 80 H (b) 0.08 H (c) 0.044 H (d) 0.065 H

(d) 0.065 H

Given I=10 A, V=80 V, R=V / I=80 / 10=8 Ω and ω=50 Hz, I=V / (8^2+X_L^2)^(1 / 2), 10=220 / (64+X_L^2)^(1 / 2), (64+X_L^2)^(1 / 2)=22, Squaring on both sides, we get 64+X_L^2=484, X_L^2=484-64=420, X_L=(420)^(1 / 2) ⇒ 2 π × ω L=(420)^(1 / 2), Series inductor on an arc lamp, L=(420)^(1 / 2) / (2 π × 50)=0.065 H

Electromagnetic Induction

Q9: A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns, giving the output power at 230 V. If the current in the primary of the transformer is 5 A, and its efficiency is 90 %, the output current would be

(a) 50 A (b) 25 A (c) 45 A (d) 20 A

(c) 45 A

Efficiency η=0.9=Ps / Pp, Vs Is=0.9 × Vp Ip, Is=(0.9 × 2300 × 5) / 230=45 A

Electromagnetic Induction

Q10: A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity ω, the maximum e.m.f. induced in the coil will be

(a) (3 / 2) nBA ω (b) nBA (c) 3 nBA ω (d) (1 / 2) nBA ω

(b) nBA

Emf induced in the coil is given by ε=BAn ω sin ω t, εmax=BA (The induced emf is maximum when sin ω t=1)

Electromagnetic Induction

Q11: A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 m/s, at right angles to the horizontal component of the earth's magnetic field of 0.3 × 10^-4 Wb/m^2. The value of the induced emf in wire is

(a) 0.3 × 10^-3 V (b) 2.5 × 10^-3 V (c) 1.5 × 10^-3 V (d) 1.1 × 10^-3 V

(c) 1.5 × 10^-3 V

The motional emf is given as ε=|v(l × B)| =5 ×(0.3 × 10^-4)(10) sin 90°=1.5 × 10^-3 V

Electromagnetic Induction

Q12: A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil

(a) increases by a factor of 27 (b) decreases by a factor of 3 (c) increases by a factor of 3 (d) decreases by a factor of 9

(c) increases by a factor of 3

Electromagnetic Induction

Q13: The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is

(a) 637.5 J (b) 540 J (c) 437.5 J (d) 740 J

(c) 437.5 J

ε=25 V, I1=10 A, I2=25 A, t=1 s, ΔE=?, ε=L(ΔI / Δt), 25=L[(25-10) / 1], L=25 / 15=(5 / 3) H, ΔE=1 / 2 L(I2^2-I1^2)=(1 / 2) ×(5 / 3) ×(25^2-10^2)=(5 / 3) × 525=437.5 J

Electromagnetic Induction

Q14: There are two long co-axial solenoids of the same length l. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self inductance of the inner coil is

(a) n2 / n1 (b) (n2 / n1)(r2^2 / r1^2) (c) (n2 / n1)(r1 / r2) (d) n1 / n2

(a) n2 / n1

The mutual inductance of the inner coil M=μ0 n1 n2 r1^2, Self inductance (L) of the inner coil is L=μ0 n1^2 r1^2, Using (1) and (2), M / L=n2 / n1

Electromagnetic Induction

Q15: A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is

(a) 2 mV (b) 6 mV (c) 1 mV (d) 12 mV

(d) 12 mV

Emf developed across the given edges, ε=Blv=0.1 × 0.02 × 6=12 × 10^-3 V=12 mV, As all the edges are parallel between the faces perpendicular to the x-axis, hence required potential difference is 12 mV

Electromagnetic Induction

Q1: If the distance between the earth and the sun were half its present value, the number of days in a year would have been

(a) 64.5 (b) 129 (c) 182.5 (d) 730

(b) 129 days

From Kepler's law, T² ∝ R³ Therefore, (T₂/T₁)² = (R₂/R₁)³ T₁ = 365 days, R₁ = R, R₂ = R/2 (T₂/365)² = (R/2/R)³ T₂² = (365)²/8 T₂² = 16,653 T₂ = 129 days

Gravitation

Q2: The mass of a spaceship is 1000 kg. It is to be launched from the earth's surface out into free space. The value of 'g' and 'R' (radius of the earth) is 10 m/s² and 6400 km respectively. The required energy for this work will be

(a) 6.4 × 10¹⁰ Joules (b) 6.4 × 10¹¹ Joules (c) 6.4 × 10⁸ Joules (d) 6.4 × 10⁰ Joules

(a) 6.4 × 10¹⁰ Joules

The energy required is given by = GMm/R = gR² × m/R (because g = GM/R²) = mgR = 1000 × 10 × 6400 × 10³ = 64 × 10⁹ J = 6.4 × 10¹⁰ J

Gravitation

Q3: An artificial satellite moving in a circular orbit around the earth has a total (K.E. + P.E.) energy E₀. Its potential energy is

(a) -E₀ (b) 1.5E₀ (c) 2E₀ (d) E₀

(c) 2E₀

Total Energy, E₀ = -GMm/2r Potential Energy, U = -GMm/r = 2E₀

Gravitation

Q4: The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is

(a) R/2 (b) R/3 (c) 2R (d) 3R

(c) 2R

Acceleration due to gravity at a height "h" is given by g' = g(R/R + h)² Here, g is the acceleration due to gravity on the surface R is the radius of the earth As g' is given as g/9, we get g/9 = g(R/R + h)² 1/3 = R/R + h h = 2R

Gravitation

Q5: A simple pendulum has a time period T₁ when on the earth's surface, and T₂ when taken to a height R above the earth's surface, where R is the radius of the earth. The value of T₂/T₁ is

(a) 1 (b) 3 (c) 4 (d) 2

(d) 2

2π√(l/g) The time period of a simple pendulum = On the surface of earth g = GM/R² At a height R above the earth g = GM/(2R)² The time period on the surface of the earth T₁ = 2π√(lR²/GM) The time period on the surface of the earth T₂ = 2π√(l(2R)²/GM) T₂/T₁ = 2

Gravitation

Q6: A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface (R = 6,400 km) will approximately be

(a) 1/2 hr (b) 1 hr (c) 2 hr (d) 4 hr

(c) 2 hr

According to Kepler's law, T² ∝ R³ Therefore, (T₂/T₁)² = (R₂/R₁)³ For a spy satellite time period is given by T₁ R₁ = 6400 km For a geostationary satellite, T₂ = 24 hours R₂ = 36,000 km (24/T₁)² = (36000/6400)³ (24/T₁)² = 178 (24/T₁) = 13.34 T₁ = 24/13.34 ≈ 2 hr

Gravitation

Q7: Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

(a) -6 Gm/r (b) -9 Gm/r (c) Zero (d) -4 Gm/r

(b) -9 Gm/r

P is the point where the field is zero, and a unit mass is placed at P. Applying Newton's law of gravitation, (Gm × 1)/x² = (G 4m × 1)/(r - x)² x²/(r - x)² = 1/4 x/(r - x) = 1/2 2x = r - x x = r/3 Potential at the point P, V = -Gm/x - G(4m)/(r - x)

Gravitation

Q8: A Binary star system consists of two stars A and B which have time periods Tₐ and Tᵦ, radii Rₐ and Rᵦ and masses Mₐ and Mᵦ. Then

(a) Tₐ > Tᵦ then Rₐ > Rᵦ (b) Tₐ > Tᵦ then Mₐ > Mᵦ (c) (Tₐ/Tᵦ)² = (Rₐ/Rᵦ)² (d) Tₐ = Tᵦ

(d) Tₐ = Tᵦ

Angular velocity of binary stars are the equal ωₐ = ωᵦ 2π/Tₐ = 2π/Tᵦ ∴ Tₐ = Tᵦ

Gravitation

Q9: Two particles of equal mass "m" go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle with respect to their centre of mass is

(a) √(Gm/R) (b) √(Gm/4R) (c) √(Gm/3R) (d) √(Gm/2R)

(b) √(Gm/4R)

Gm²/4R² = mv²/R V = √(Gm/4R)

Gravitation

Q10: A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass "m" is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

(a) 1/2 mV² (b) mV² (c) 3/2 mV² (d) 2 mV²

(b) mV²

Kinetic energy at the time of ejection = 1/2 mvₑ² Vₑ is the escape velocity = √2 × orbital velocity Vₑ is the escape velocity = √2 v Kinetic energy at the time of ejection = 1/2 m(√2 v)² = mv²

Gravitation

Q11: Three particles, each of mass m, are situated at the vertices of an equilateral triangle of side length a. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time period of the circular motion. (F = Gm₁m₂/r²)

V = √(Gm/a) and T = 2π √(a³/3 Gm)

ABC is an equilateral triangle of side a Each particle moves in a circle of radius r AD² = a² - a²/4, r = 2/3 AD r = 2/3 × √(a² - a²/4) r = a √3---(1) To find v Let v = Initial velocity given to each particle. For circular motion, centripetal force should be provided. It is provided by the gravitational force between two masses. Let F denote this force. F = Gm²/a² Resultant force = √(F² + F² + 2 F² cos 60°) Resultant force = √3 F Resultant force = Centripetal force √3 F = mv²/r V² = √3 F r/m = √3/m × (Gm²/a²) (a/√3) = Gm/a V = √(Gm/a) To find time period of circular motion(T) T = 2πr/V = 2π(a/√3) × √(a/ Gm) = 2π √(a³/3 Gm) therefore, T = 2π √(a³/3 Gm)

Gravitation

Q12: What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?

(a) 5 GmM/6 R (b) 2 GmM/3 R (c) GmM/2 R (d) GmM/3 R

(a) 5 GmM/6 R

Energy of the satellite on the surface of the Earth E₁ = -GMm/R Energy at a distance 2R is given by E₂ = -GMm/3R + 1/2 mv₀² E₂ = -GMm/3R + 1/2 m[GM/3R] E₂ = -GmM/6R E₂ - E₁ = (-GmM/6R) - (-GMm/R) = 5GmM/6R

Gravitation

Q13: Two satellites S1 and S2 revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are 1 hour and 8 hours, respectively. The radius of the orbit of S1 is 10⁴ km. When S2 is closest to S1, find the angular speed of S2 as actually observed by an astronaut in S1

Angular speed = 7π/8

Angular velocity ω = 2π/T Here T = 1 hour for S₁, ω₁ = 2π rad/hr Similarly, for S₂, ω₂ = 2π Given that they are rotating in the same sense. So, relative angular velocity = 2π - 2π/8 = 7π/8

Gravitation

Q14: Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence

(a) There will be no change in weight anywhere on the earth (b) Weight of the object, everywhere on the earth, will increase (c) Except at poles, weight of the object on the earth will decrease (d) Weight of the object, everywhere on the earth, will decrease.

(c) Except at poles, weight of the object on the earth will decrease

The effect of rotation of earth on acceleration due to gravity is given by g' = g - ω²R cos 2Φ Where Φ is latitude. There will be no change in gravity at poles as Φ = 90°, at all points as ω increases g' will decrease.

Gravitation

Q15: A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nᵗʰ power of R. If the period of rotation of the particle is T, then

(a) T ∝ R³/₂ (b) T ∝ R(m 2)+1 (c) T ∝ R(n+1)/2 (d) T ∝ Rⁿ/2

(c) T ∝ R(n+1)/2

According to the question, central force is given by Fᶜ ∝ 1/Rⁿ Fᶜ = k(1/Rⁿ) mω²R = k(1/Rⁿ) m(2π)²/T² = k(1/Rⁿ+¹) Or T² ∝ Rⁿ+¹ T ∝ R(n+1)/2

Gravitation

Q1: A solid sphere of radius $R$ acquires a terminal velocity $v_{1}$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into 27 identical spheres. If each of these acquires a terminal velocity $v_{2}$, when falling through the same fluid, the ratio ( $\left.v_{1} / v_{2} ight)$ equals

(a) 9 (b) $1 / 27$ (c) $1 / 9$ (d) 27

(a) 9

$27 \mathrm{x}(4 / 3) \pi \mathrm{r}^{3}=(4 / 3) \pi \mathrm{R}^{3}$ $\operatorname{Or} \mathrm{r}=\mathrm{R} / 3$ Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{3}$ Therefore, $\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight)=\left(\mathrm{R}^{2} / \mathrm{r}^{2} ight)$ $\mathrm{v}_{1} \mathrm{v}_{2}=[\mathrm{R} /(\mathrm{R} / 3)]^{2}=9$ $\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight)=9$

Fluid Mechanics

Q2: Spherical balls of radius $R$ are falling in a viscous fluid of viscosity with a velocity $v$. The retarding viscous force acting on the spherical ball is

(a) directly proportional to $\mathrm{R}$ but inversely proportional to $\mathrm{v}$ (b) directly proportional to both radius $\mathrm{R}$ and velocity $\mathrm{v}$ (c) inversely proportional to both radius $\mathrm{R}$ and velocity $\mathrm{v}$ (d) inversely proportional to $\mathrm{R}$ but directly proportional to velocity $\mathrm{v}$

(b) directly proportional to both radius $R$ and velocity $v$

Retarding viscous force $=6 \pi \eta R v$ obviously option (b) holds goods

Fluid Mechanics

Q3: A long cylindrical vessel is half-filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is $5 \mathrm{~cm}$ and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in $\mathrm{cm}$, will be

(a) 0.4 (b) 2.0 (c) 0.1 (d) 1.2

(b) 2.0

The linear speed of the liquid at the sides is $\mathrm{r} \omega$. So, the difference in height is given as follows $2 \mathrm{gh}=\omega^{2} \mathrm{r}^{2}$ $\mathrm{h}=\omega^{2} \mathrm{r}^{2} / 2 \mathrm{~g}$ here $\omega=2 \pi \mathrm{f}$ Therefore, $\mathrm{h}=\left[(2 imes 2 \pi)^{2}\left(5 imes 10^{-2} ight)^{2} ight] /(2 imes 10)=2 \mathrm{~cm}$

Fluid Mechanics

Q4: Water is flowing continuously from a tap having an internal diameter $8 imes 10^{-3} \mathrm{~m}$. The water velocity as it leaves the tap is $0.4 \mathrm{~ms}^{-1}$. The diameter of the water stream at a distance $2 imes 10^{-1} \mathrm{~m}$ below the tap is close to

(a) $5.0 imes 10^{-3} \mathrm{~m}$ (b) $7.5 imes 10^{-3} \mathrm{~m}$ (c) $9.6 imes 10^{-3} \mathrm{~m}$ (d) $3.6 imes 10^{-3} \mathrm{~m}$

(d) $3.6 imes 10^{-3} \mathrm{~m}$

Here, $\mathrm{d}_{1}=8 imes 10^{-3} \mathrm{~m}$ $\mathrm{v}_{1}=0.4 \mathrm{~m} \mathrm{~s}^{-1}, \mathrm{~h}=0.2 \mathrm{~m}$ According to equation of motion, $$ \begin{aligned} & \quad v_{2}=\sqrt{v_{1}^{2}+2 g h}=\sqrt{(0.4)^{2}+2+10 imes 0.2} \n& =2 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$ According to equation of continuity $\mathrm{a}_{1} \mathrm{v}_{1}=\mathrm{a}_{2} \mathrm{v}_{2}$ According to equation of continuity $\mathrm{a}_{1} \mathrm{v}_{1}=\mathrm{a}_{2} \mathrm{v}_{2}$ $\left(\pi \mathrm{D}_{1}{ }^{2} / 4 ight) \mathrm{v}_{1}=\left(\pi \mathrm{D}_{2}{ }^{2} / 4 ight) \mathrm{v}_{2}$ $\mathrm{D}_{2}{ }^{2}=\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight) \mathrm{D}_{1}{ }^{2}$ $\mathrm{D}_{2}=\left[\sqrt{ }\left(\mathrm{v}_{1} / \mathrm{v}_{2} ight) ight] \mathrm{D}_{1}$ $=[\sqrt{ }(0.4 / 2)] imes 8 imes 10^{-3} \mathrm{~m}$ $\mathrm{D}_{2}=3.6 imes 10^{-3} \mathrm{~m}$

Fluid Mechanics

Q5: A $20 \mathrm{~cm}$ long capillary tube is dipped in water. The water rises up to $8 \mathrm{~cm}$. If the entire arrangement is put in a freely falling elevator the length of the water column in the capillary tube will be

(a) $4 \mathrm{~cm}$ (b) $20 \mathrm{~cm}$ (c) $8 \mathrm{~cm}$ (d) $10 \mathrm{~cm}$

(b) $20 \mathrm{~cm}$

In a freely falling elevator, $\mathrm{g}=0$ Water will rise to the full length i.e., $20 \mathrm{~cm}$ to tube

Fluid Mechanics

Q6: A spherical solid ball of volume $V$ is made of a material of density $ ho_{1}$. It is falling through a liquid of density $ ho_{2}\left( ho_{2}<1 ight)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$, i.e., $F_{ ext {viscous }}=-\mathbf{k v}^{2}(k>0)$. The terminal speed of the ball is

(a) $\operatorname{Vg}\left( ho_{1}- ho_{2} ight)$ (b) \[ \sqrt{ \frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}} \] (c) $\operatorname{Vg} ho_{1} / \mathrm{k}$ (d) $\sqrt{\frac{V g ho_{1}}{k}}$

(b) \[ \sqrt{ \frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}} \]

The forces acting on the solid ball when it is falling through a liquid is "mg" downwards, thrust by Archimedes principle upwards and the force due to the force of friction also acting upwards. The viscous force rapidly increases with velocity, attaining a maximum when the ball reaches the terminal velocity. Then the acceleration is zero $\mathrm{mg}-\mathrm{V} ho_{2} \mathrm{~g}-\mathrm{kv}_{\mathrm{t}}^{2}=$ ma where $\mathrm{V}$ is volume, $\mathrm{v}_{\mathrm{t}}$ is the terminal velocity When the ball is moving with terminal velocity, $\mathrm{a}=0$ Therefore $\mathrm{V} ho_{1} \mathrm{~g}-\mathrm{V} ho_{2} \mathrm{~g}-\mathrm{kv}_{\mathrm{t}}^{2}=0$ $\mathrm{v}_{\mathrm{t}}=\sqrt{\frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}}$

Fluid Mechanics

Q7: Water flows into a large tank with a flat bottom at the rate of $10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1}$. Water is also leaking out of a hole of area $1 \mathrm{~cm}^{2}$ at its button. If the height of the water in the tank remains steady, then this height is

(a) $5 \mathrm{~cm}$ (b) $7 \mathrm{~cm}$ (c) $4 \mathrm{~cm}$ (d) $9 \mathrm{~cm}$

(a) $5 \mathrm{~cm}$

Since the height of the water column is constant Water inflow rate $\left(\mathrm{Q}_{ ext {in }} ight)=$ Water outflow rate $\left(\mathrm{Q}_{ ext {out }} ight)$ $\mathrm{Q}_{ ext {in }}=10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1}$ $\mathrm{Q}_{ ext {out }}=10^{-4} \mathrm{x} \sqrt{ }(2 \mathrm{gh})$ $10^{-4}=10^{-4} \mathrm{x} \sqrt{ } 20 \mathrm{xh}$ $\mathrm{h}=(1 / 20) \mathrm{m}=5 \mathrm{~cm}$

Fluid Mechanics

Q8: A submarine experiences a pressure of $5.05 imes 10^{6} \mathrm{~Pa}$ at depth of $d_{1}$ in a sea. When it goes further to a depth of $d_{2}$, it experiences a pressure of $8.08 imes 10^{6} \mathrm{~Pa}$. Then $d_{1}-d_{2}$ is approximately (density of water $=10^{3}$ $\mathrm{ms}^{-2}$ and acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )

(a) $300 \mathrm{~m}$ (b) $400 \mathrm{~m}$ (c) $600 \mathrm{~m}$ (d) $500 \mathrm{~m}$

(a) $300 \mathrm{~m}$

$\mathrm{P}_{1}=\mathrm{P}_{0}+ ho \mathrm{gd}_{1}$ $\mathrm{P}_{2}=\mathrm{P}_{0}+ ho \mathrm{gd}_{2}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}= ho g \Delta \mathrm{d}$ $\left(8.08 imes 10^{6}-5.05 imes 10^{6} ight)=10^{3} imes 10 imes \Delta \mathrm{d}$ $3.03 imes 10^{6}=10^{3} imes 10 imes \Delta \mathrm{d}$ $\Delta \mathrm{d}=303 \mathrm{~m} pprox 300 \mathrm{~m}$

Fluid Mechanics

Q9: Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is $5 \mathrm{~cm}$, the Reynolds number for the flow is of the order (density of water $=1000 \mathrm{~kg} / \mathrm{m}^{3}$, coefficient of viscosity of water $=1 \mathrm{mPa} \mathrm{s}$ )

(a) $10^{3}$ (b) $10^{4}$ (c) $10^{2}$ (d) $10^{6}$

(b) $10^{4}$

Reynolds number $= ho v d / \eta$ Volume flow rate $=\mathrm{vx} \pi \mathrm{r}^{2}$ $\mathrm{v}=\left(100 imes 10^{-3} / 60 ight) imes\left(1 / \pi imes 25 imes 10^{-4} ight)$ $\mathrm{v}=(2 / 3 \pi) \mathrm{m} / \mathrm{s}$ Reynolds number $=\left\{\left(10^{3} imes 2 imes 10 imes 10^{-2} ight) /\left(10^{-3} imes 3 \pi\right)\right\}$ $\simeq 2 imes 10^{4}$ Order of $10^{4}$

Fluid Mechanics

Q10: The top of a water tank is open to the air and its water level is maintained. It is giving out $0.74 \mathrm{~m}^{3}$ water per minute through a circular opening of $2 \mathrm{~cm}$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to

(a) $6.0 \mathrm{~m}$ (b) $4.8 \mathrm{~m}$ (c) $9.6 \mathrm{~m}$ (d) $2.9 \mathrm{~m}$

(b) $4.8 \mathrm{~m}$

Here, volumetric flow rate $=(0.74 / 60)=\pi r^{2} v=\left(\pi imes 4 imes 10^{-4} ight) imes \sqrt{ } 2$ gh $\Rightarrow \sqrt{ } 2 \mathrm{gh}=[(74 imes 100) / 240 \pi)]$ $\Rightarrow \sqrt{ } 2 \mathrm{gh}=740 / 24 \pi$ $2 \mathrm{gh}=(740 / 24 \pi)^{2}$ $\mathrm{h}=[(740 imes 740) / 24 imes 24 imes 10)]\left(\right.$ since $\left.\pi^{2}=10\right)$ $\mathrm{h} pprox 4.8 \mathrm{~m}$

Fluid Mechanics

Q1: During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts

(a) in the backward direction on the front wheel and in the forward direction on the rear wheel (b) in the forward direction on the front wheel and in the backward direction on the rear wheel (c) in the backward direction on both, the front and the rear wheels (d) in the forward direction on both, the front and the rear wheels.

Answer: (a) Due to peddling, the point of contact of the rear wheel has a tendency to move backwards. So frictional force opposes the backwards tendency i.e., the frictional force acts in the forward direction. But the back wheel accelerates the front wheel in the forward direction. To oppose this frictional force acts in the backward direction on the front wheel.

Friction

Q2: A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5 , the magnitude of the frictional force acting on the block is

(a) 2.5 N (b) 0.98 N (c) 4.9 N (d) 0.49 N

Answer: (a) 2.5 N

Solution: Consider the forces, acting on the block in the vertical direction Force of friction f=μR f=0.5×5 f=2.5 N

Friction

Q3: A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2 . The weight of the block is-

(a) 20 N (b) 50 N (c) 100 N (d) 2 N

Answer: (d) 2N

Solution: Frictional force balances the weight of the body Frictional force f=μN=mg f=0.2X 10=2 N Therefore, weight of the block mg=2 N

Friction

Q4: A marble block of mass 2 kg lying on ice when given a velocity of 6 m / s is stopped by friction in 10 s. Then the coefficient of friction is (consider g=10 m / s)

(a) 0.02 (b) 0.03 (c) 0.06 (d) 0.01

Answer: (c) 0.06

Solution: u=6 m / s v=0 t=10 s a=-f / m=-μmg / m=-μg=-10 μ Substituting values in v=u+at 0=6-10 μ× 10 Therefore, μ=0.06

Friction

Q5: A block P of mass m is placed on a horizontal frictionless plane. The second block of the same mass m is placed on it and is connected to a spring of spring constant k. The two blocks are pulled by a distance A. Block Q oscillates without slipping. What is the maximum value of the frictional force between the two block.

(a)kA/2 (b) kA (c) μs mg (d) zero

Answer: (a) fm=k A / 2

Solution: Block Q oscillates but does not slip on P. It means that acceleration is the same for Q and P both. There is a force of friction between the two blocks while the horizontal plane is frictionless. The spring is connected to the upper block. The (P-Q) system oscillates with angular frequency ω. The spring is stretched by A. Therefore, ω=√k / 2 m Therefore, Maximum acceleration in SHM=ω^2 A a_m=kA / 2 m Now consider the lower block. Let the maximum force of friction =f_m f_m=ma_m or f_m=mx(kA / 2 m) f_m=kA / 2

Friction

Q6: Statement 1: A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in mechanical energy in the second situation is smaller than that in the first situation.

Statement-2: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination. (a) Statement-1 and 2 are true and statement-2 is a correct explanation for statement-1 (b) Statement-1 and 2 are true and statement-2 is not a correct explanation for statement-1 (c) Statement-1 is true, statement-2 is false (d) Statement-1 is false, statement-2 is true.

Answer: (c) Statement-1 is true, statement-2 is false

Friction

Q7: A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8 . If the frictional force on the block is 10 N, the mass of the block (in kg) is : (taken g=10 m / s² )

(a) 2.0 (b) 4.0 (c) 1.6 (d) 2.5

Answer: (a) 2

Solution: μ=0.8 f=10 N mg sin 30=10 mx 10×(1 / 2)=10 m=2 kg

Friction

Q8: A smooth block is released at rest on a 45° incline and then slides a distance d. The time taken to slide is n times as much to slide on a rough incline than on a smooth incline. The coefficient of friction is-

(a) μ k=1-1 / n² (b) μ k=√(1-1 / n²) (c) μ s=1-1 / n² (d) μ s=√(1-1 / n²)

Answer: (a) μ k=1-1 / n²

Solution: On a smooth plane d=1 / 2(g sin θ) t_1 t_1=√(2 d / g sin θ) On a rough surface d=1 / 2(g sin θ-μ g cos θ) t_2 t_2=√(2 d / g sin θ-μ g cos θ) From the question, we know t_2=nt_1 √(2 d / g sin θ-μ g cos θ) n√(2 d / g sin θ) n=1 / √(1-μ_k) (since cos 45°=sin 45° 1 / √2) n²=1 / 1-μ_k 1-μ_k=1 / n² μ_k=1-1 / n²

Friction

Q9: The upper half of an inclined plane with inclination Φ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by

(a) 2 sin Φ (b) 2 cos Φ (c) 2 tan Φ (d) tan Φ

Answer: (c) 2 tan Φ

Solution: Suppose the length of the plane is L. When the block slides down the plane, the increase in the K.E will be equal to the decrease in the P.E Work done, W= Change in K.E(ΔK)=(1 / 2) mu²-(1 / 2) mv²=0 Work done by friction (W_f)+ Work done by gravity (W_g)=0 -μ g cos Φ(L / 2)+mgLsin Φ=0 (μ / 2) cos Φ=sin Φ μ=2 tan Φ

Friction

Q10: Consider a car moving on a straight road with a speed of 100 m / s. The distance at which a car can be stopped is : (μ_k=0.5)

(a) 800 m (b) 1000 m (c) 100 m (d) 400 m

Answer: (b) 1000

Solution: When the car is stopped by friction then its retarding force is ma=μ R ma=μ mg a=μ g Consider the equation v²=u²-2 as (the car is retarding so a is negative) 0=u²-2 as 2as =u² s=u² / 2 a s=u² / 2 μ g s=(100)² / 2× 0.5× 10 s=1000 m

Friction

Q11: The minimum force required to start pushing a body up a rough (frictional coefficient μ ) inclined plane F₁ while the minimum force needed to prevent it from sliding down is F₂. If the inclined plane makes an angle θ from the horizontal such that tan θ=2 μ then the ratio F₁ / F₂ is

(a) 4 (b) 1 (c) 2 (d) 3

Answer: (d) 3

Solution: We have to work against the sliding force due to gravity and frictional force to push upwards. Therefore, the force F₁=mg sin θ+μ mg cos θ. To stop the body from sliding work is done against sliding force but frictional force stops the body from sliding. Therefore, the force F₂=mg sin θ-μ mg cos θ. F₁ / F₂=(mg sin θ+μ mg cos θ) /(mg sin θ-μ mg cos θ) F₁ / F₂=(tan θ+μ) /(tan θ-μ) (since tan θ=2 μ ) F₁ / F₂=(2 μ+μ) /(2 μ-μ) F₁ / F₂=3

Friction

Q12: A body of mass m=10⁻² kg is moving in a medium and experiences a frictional force F=-kv². Its initial speed is v₀=10 ms⁻¹. If, after 10 s, its energy is 1 / 8 mv₀², the value of k will be

(a) 10⁻³ kg m⁻¹ (b) 10⁻³ kg s⁻¹ (c) 10⁻⁴ kg m⁻¹ (d) 10⁻¹ kg m⁻¹ s⁻¹

Answer: (c) 10⁻⁴ kg m⁻¹

Solution: Mass m=10² Kg Initial velocity v₀=10 ms⁻¹ Time t=10 s Frictional Force F=-kv² 1 / 2 mv_f²=1 / 8 mv₀² v_f=v₀ / 2 v_f=10 / 2=5 m / s Now, the force is (10⁻²) dv / dt=-kv² ∫_{10}^{5} (dv / v²)=-100 k ∫_{0}^{10} (dt) 1 / 5=1 / 10=100 kx 10 k=10⁻⁴

Friction

Q13: A block of mass m is placed on a surface with a vertical cross-section given by y=x³ / 6. If the coefficient of friction is 0.5 , the maximum height above the ground at which the block can be placed without slipping is

(a) 1 / 3 m (b) 1 / 2 m (c) 1 / 6 m (d) 2 / 3 m

Answer: (c) 1 / 6

Solution: For equilibrium under limiting friction mg sin θ=μ mg cos θ tan θ=μ The equation of the surface is given by y=x³ / 6 Slope dy / dx=x² / 2 From (1) and (2) we get μ=x² / 2=0.5 x=1 So, y=1 / 6

Friction

Q2: The ratio of work done by an ideal monoatomic gas to the heat supplied to it in an isobaric process is

(a) 2 / 5 (b) 3 / 2 (c) 3 / 5 (d) 2 / 3

(a) 2 / 5

For an ideal gas in an isobaric process, Heat supplied, Q=nCp ΔT Work done, W=P ΔV=nR ΔT Required Ratio =W / Q Required Ratio =(nR ΔT)/(nCp ΔT) R / Cp=R/(γR / γ-1) [since γ=5 / 3, for a monoatomic gas ] =2 / 5

Thermal Expansion

Q3: The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7°C. The gas is (R=8.3 J mol−1 K−1)

(a) monoatomic (b) diatomic (c) triatomic (d) a mixture of monoatomic and diatomic

(b) diatomic

Increase in temperature, ΔT=T2−T1=7 K (Since it is change in temperature so its value will be same in Kelvin and degree celsius) Work done on the system, W=146 KJ=146 × 1000 J Number of moles of gas, n=1000 Work done in an adiabatic process, W=nR(T2−T1)/γ-1 put the values, 146000=1000 × 8.3 × 7 /(γ-1) γ-1=0.39 γ≈1.4 Hence the gas is diatomic

Thermal Expansion

Q4: Two moles of an ideal monatomic gas occupies a volume V at 27°C. The gas expands adiabatically to a volume 2 V. Calculate (i) the final temperature of the gas and (ii) change in its internal energy.

(a) (i) 198 K (ii) 2.7 kJ (b) (i) 195 K (ii) 2.7 kJ (c) (i) 189 K (ii) 2.7 kJ (d) (i) 195 K (ii) 2.7 kJ

(c) (i) 189 K (ii) 2.7 kJ

For an adiabatic process, PV^γ=constant (nRT/V)V^γ=constant or TV^(γ-1)=constant T1 V1^(γ-1)=T2 V2^(γ-1) T2=T1[V1 / V2]^(γ-1) Here, T1=27°C=300 K, V1=V, V2=2 V, γ=5 / 3 T2=300(V / 2 V)(5 / 3-1)=300(1 / 2)^(2 / 3)=189 K Change in internal energy, ΔU=nCv ΔT ΔU=n[(f / 2)R](T2−T1)=2 ×(3 / 2) ×(25 / 3)(189-300) ΔU=-2.7 kJ

Thermal Expansion

Q6: The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 1000°C is (For steel Young's modulus is 2 × 10^11 Nm−2 and coefficient of thermal expansion is 1.1 × 10−5 K−1)

(a) 2.2 × 10^7 Pa (b) 2.2 × 10^6 Pa (c) 2.2 × 10^8 Pa (d) 2.2 × 10^9 Pa

(c) 2.2 × 10^8 Pa

Given, ΔT=100°C, Y=2 × 10^11 N m−2 α=1.1 × 10−5 K−1 Young's modulus, Y= stress/strain =(F / A) /ΔL / L Therefore, Y=P /ΔL / L Y=P / α ΔT [ since ΔL=L α ΔT] Thermal stress in a rod is the pressure due to the thermal strain. Required pressure P=Y α ΔT=2 × 10^11 × 1.1 × 10−5 × 100=2.2 × 10^8 Pa

Thermal Expansion

Q7: Two wires are made of the same material and have the same volume. However, wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount?

(a) 6 F (b) 9 F (c) F (d) 4 F

(d) 9 F

For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire 11 is A and that of 12 is 3 A. V=V1=V2 V=A × 11=3 A × 12 12=11 / 3 Y=(F / A) /Δ1 / l ⇒F1=YA(Δ11 / 11) F2=Y 3 A(Δl2 / l2) Given Δl1=Δl2=Δx (for the same extension) F2=Y 3 A(Δx / 11 / 3)=9(YA Δx / 11)=9 F1 or 9 F

Thermal Expansion

Q8: A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to

(a) 9 F / (π r^2 YT) (b) F / (3 π r^2 YT) (c) 6 F / (π r^2 YT) (d) 3 F / (π r^2 YT)

(d) 3 F / (π r^2 YT)

Given, Length =L Longitudinal force =F Radius =r Young's modulus =Y Temperature =T Since length remains same (Stress)Compressive =( Stress ) Thermal F / A=Y α T ⇒F / π r^2=Y α T (where α= coeficient of linear expansion) α=F / π r 2 YT Coefficient of volume expansion, γ=3 α γ=3 F / π r^2 YT

Thermal Expansion

Q9: During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio CP / CV for the gas is

(a) 4 / 3 (b) 2 (c) 5 / 3 (d) 3 / 2

(d) 3 / 2

In an adiabatic process, T^γ= (constant) P^γ-1 T^γ / γ-1= (constant)P Given T^3= (constant) P γ / γ-1=3 ⇒ 3 γ-3=γ ⇒ γ=3 / 2

Thermal Expansion

Q10: An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

(a) 3 / P α K (b) P / 3 α K (c) 3 α / PK (d) 3 PK α

(b) P / 3 α K

The bulk modulus of the gas is given by K=-P /ΔV / V 0 (Here negative sign indicates the decrease in volume with pressure) ΔV / V0=P / K [in magnitude] Also, V=V0(1+γ ΔT) or ΔV / V0=γ ΔT Comparing eq. (1) and (2), we get P / K=γ ΔT ΔT=P / 3 α K (since γ=3 α )

Thermal Expansion

Q11: Steel wire of length ' L ' at 40°C is suspended from the ceiling and then a mass ' m ' is hung from its free end. The wire is cooled down from 40°C to 30°C to regain its original length ' L '. The coefficient of linear thermal expansion of the steel is 10−5 /°C, Young's modulus of steel is 10^11 N / m^2 and radius of the wire is 1 mm. Assume that L > diameter of the wire. Then the value of ' m ' in kg is nearly

3

E=(F / A) /ΔL / L =(F × L)/(A × ΔL) ⇒(ΔL / L)=F / AE Since, (ΔL / L)=α Δt α Δt=F / AE α Δt=mg / AE m=[(α Δt) / g] AE m=3.14 ≈ 3

Thermal Expansion

Q12: A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

(a) 25 J (b) 35 J (c) 30 J (d) 40 J

(b) 35 J

The value of Cp for a diatomic gas is Cp=(7 / 2) R-(1) It is given that work done during expansion at constant pressure is dW=10 J=nR ΔT Since the gas undergoes an isobaric process ⇒ ΔQ=n(7 / 2) R ΔT=7 / 2(nR ΔT)=(7 / 2)(10)=35 J

Thermal Expansion