Rishabhiiitd/PhysicsQA huggingface dataset:
Topic: 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
Options: (a) n2 / n1
(b) (n2 / n1)(r2^2 / r1^2)
(c) (n2 / n1)(r1 / r2)
(d) n1 / n2
Answer: (a) n2 / n1
Solution: 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
Topic: 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
Options: (a) 2 mV
(b) 6 mV
(c) 1 mV
(d) 12 mV
Answer: (d) 12 mV
Solution: 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
Topic: Gravitation
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
Options: (a) 64.5
(b) 129
(c) 182.5
(d) 730
Answer: (b) 129 days
Solution: 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
Topic: 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
Options: (a) 6.4 × 10¹⁰ Joules
(b) 6.4 × 10¹¹ Joules
(c) 6.4 × 10⁸ Joules
(d) 6.4 × 10⁰ Joules
Answer: (a) 6.4 × 10¹⁰ Joules
Solution: 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
Topic: 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
Options: (a) -E₀
(b) 1.5E₀
(c) 2E₀
(d) E₀
Answer: (c) 2E₀
Solution: Total Energy, E₀ = -GMm/2r
Potential Energy, U = -GMm/r = 2E₀
Topic: 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
Options: (a) R/2
(b) R/3
(c) 2R
(d) 3R
Answer: (c) 2R
Solution: 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
Topic: 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
Options: (a) 1
(b) 3
(c) 4
(d) 2
Answer: (d) 2
Solution: 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
Topic: 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
Options: (a) 1/2 hr
(b) 1 hr
(c) 2 hr
(d) 4 hr
Answer: (c) 2 hr
Solution: 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
Topic: 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
Options: (a) -6 Gm/r
(b) -9 Gm/r
(c) Zero
(d) -4 Gm/r
Answer: (b) -9 Gm/r
Solution: 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)
Topic: 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
Options: (a) Tₐ > Tᵦ then Rₐ > Rᵦ
(b) Tₐ > Tᵦ then Mₐ > Mᵦ
(c) (Tₐ/Tᵦ)² = (Rₐ/Rᵦ)²
(d) Tₐ = Tᵦ
Answer: (d) Tₐ = Tᵦ
Solution: Angular velocity of binary stars are the equal ωₐ = ωᵦ
2π/Tₐ = 2π/Tᵦ
∴ Tₐ = Tᵦ
Topic: 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
Options: (a) √(Gm/R)
(b) √(Gm/4R)
(c) √(Gm/3R)
(d) √(Gm/2R)
Answer: (b) √(Gm/4R)
Solution: Gm²/4R² = mv²/R
V = √(Gm/4R)
Topic: 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
Options: (a) 1/2 mV²
(b) mV²
(c) 3/2 mV²
(d) 2 mV²
Answer: (b) mV²
Solution: 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
Topic: 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)
Answer: V = √(Gm/a) and T = 2π √(a³/3 Gm)
Solution: 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)
Topic: 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?
Options: (a) 5 GmM/6 R
(b) 2 GmM/3 R
(c) GmM/2 R
(d) GmM/3 R
Answer: (a) 5 GmM/6 R
Solution: 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
Topic: 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
Answer: Angular speed = 7π/8
Solution: 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
Topic: Gravitation
Q14: Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence
Options: (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.
Answer: (c) Except at poles, weight of the object on the earth will decrease
Solution: 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.
Topic: 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
Options: (a) T ∝ R³/₂
(b) T ∝ R(m 2)+1
(c) T ∝ R(n+1)/2
(d) T ∝ Rⁿ/2
Answer: (c) T ∝ R(n+1)/2
Solution: 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
Topic: Fluid Mechanics
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
Options: (a) 9
(b) $1 / 27$
(c) $1 / 9$
(d) 27
Answer: (a) 9
Solution: $27 \mathrm{x}(4 / 3) \pi \mathrm{r}^{3}=(4 / 3) \pi \mathrm{R}^{3}$
$\operatorname{Or} \mathrm{r}=\mathrm{R} / 3$
Terminal velocity, $\mathrm{v} \propto \mathrm{r}^{3}$
Therefore, $\left(\mathrm{v}_{1} / \mathrm{v}_{2}ight)=\left(\mathrm{R}^{2} / \mathrm{r}^{2}ight)$
$\mathrm{v}_{1} \mathrm{v}_{2}=[\mathrm{R} /(\mathrm{R} / 3)]^{2}=9$
$\left(\mathrm{v}_{1} / \mathrm{v}_{2}ight)=9$
Topic: 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
Options: (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 $\mathrm{v}$
Answer: (b) directly proportional to both radius $R$ and velocity $v$
Solution: Retarding viscous force $=6 \pi \eta R v$
obviously option (b) holds goods
Topic: 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
Options: (a) 0.4
(b) 2.0
(c) 0.1
(d) 1.2
Answer: (b) 2.0
Solution: 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, $\mathrm{h}=\left[(2 	imes 2 \pi)^{2}\left(5 	imes 10^{-2}ight)^{2}ight] /(2 	imes 10)=2 \mathrm{~cm}$
Topic: 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
Options: (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) $3.6 	imes 10^{-3} \mathrm{~m}$
Answer: (d) $3.6 	imes 10^{-3} \mathrm{~m}$
Solution: 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 $\mathrm{a}_{1} \mathrm{v}_{1}=\mathrm{a}_{2} \mathrm{v}_{2}$
$\left(\pi \mathrm{D}_{1}{ }^{2} / 4ight) \mathrm{v}_{1}=\left(\pi \mathrm{D}_{2}{ }^{2} / 4ight) \mathrm{v}_{2}$
$\mathrm{D}_{2}{ }^{2}=\left(\mathrm{v}_{1} / \mathrm{v}_{2}ight) \mathrm{D}_{1}{ }^{2}$
$\mathrm{D}_{2}=\left[\sqrt{ }\left(\mathrm{v}_{1} / \mathrm{v}_{2}ight)ight] \mathrm{D}_{1}$
$=[\sqrt{ }(0.4 / 2)] 	imes 8 	imes 10^{-3} \mathrm{~m}$
$\mathrm{D}_{2}=3.6 	imes 10^{-3} \mathrm{~m}$
Topic: 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
Options: (a) $4 \mathrm{~cm}$
(b) $20 \mathrm{~cm}$
(c) $8 \mathrm{~cm}$
(d) $10 \mathrm{~cm}$
Answer: (b) $20 \mathrm{~cm}$
Solution: In a freely falling elevator, $\mathrm{g}=0$ Water will rise to the full length i.e., $20 \mathrm{~cm}$ to tube
Topic: 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
Options: (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) $\sqrt{\frac{V g ho_{1}}{k}}$
Answer: (b) \[ \sqrt{ \frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}} \]
Solution: 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 $\mathrm{V} ho_{1} \mathrm{~g}-\mathrm{V} ho_{2} \mathrm{~g}-\mathrm{kv}_{\mathrm{t}}^{2}=0$
$\mathrm{v}_{\mathrm{t}}=\sqrt{\frac{V g\left(\rho_{1}-\rho_{2}\right)}{k}}$
Topic: 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
Options: (a) $5 \mathrm{~cm}$
(b) $7 \mathrm{~cm}$
(c) $4 \mathrm{~cm}$
(d) $9 \mathrm{~cm}$
Answer: (a) $5 \mathrm{~cm}$
Solution: Since the height of the water column is constant
Water inflow rate $\left(\mathrm{Q}_{	ext {in }}ight)=$ Water outflow rate $\left(\mathrm{Q}_{	ext {out }}ight)$
$\mathrm{Q}_{	ext {in }}=10^{-4} \mathrm{~m}^{3} \mathrm{~s}^{-1}$
$\mathrm{Q}_{	ext {out }}=10^{-4} \mathrm{x} \sqrt{ }(2 \mathrm{gh})$
$10^{-4}=10^{-4} \mathrm{x} \sqrt{ } 20 \mathrm{xh}$
$\mathrm{h}=(1 / 20) \mathrm{m}=5 \mathrm{~cm}$
Topic: 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 $=10 \mathrm{~ms}^{-2}$ )
Options: (a) $300 \mathrm{~m}$
(b) $400 \mathrm{~m}$
(c) $600 \mathrm{~m}$
(d) $500 \mathrm{~m}$
Answer: (a) $300 \mathrm{~m}$
Solution: $\mathrm{P}_{1}=\mathrm{P}_{0}+ho \mathrm{gd}_{1}$
$\mathrm{P}_{2}=\mathrm{P}_{0}+ho \mathrm{gd}_{2}$
$\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}=ho g \Delta \mathrm{d}$
$\left(8.08 	imes 10^{6}-5.05 	imes 10^{6}ight)=10^{3} 	imes 10 	imes \Delta \mathrm{d}$
$3.03 	imes 10^{6}=10^{3} 	imes 10 	imes \Delta \mathrm{d}$
$\Delta \mathrm{d}=303 \mathrm{~m} pprox 300 \mathrm{~m}$
Topic: 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 $=1 \mathrm{mPa} \mathrm{s}$ )
Options: (a) $10^{3}$
(b) $10^{4}$
(c) $10^{2}$
(d) $10^{6}$
Answer: (b) $10^{4}$
Solution: 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 $10^{4}$
Topic: 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
Options: (a) $6.0 \mathrm{~m}$
(b) $4.8 \mathrm{~m}$
(c) $9.6 \mathrm{~m}$
(d) $2.9 \mathrm{~m}$
Answer: (b) $4.8 \mathrm{~m}$
Solution: 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 $\left.\pi^{2}=10\right)$
$\mathrm{h} pprox 4.8 \mathrm{~m}$
Topic: Friction
Q1: During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts
Options: (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: 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.
Topic: 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
Options: (a) 2.5 N
(b) 0.98 N
(c) 4.9 N
(d) 0.49 N
Answer: Answer: (a) 2.5 N
Solution: 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
Topic: 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-
Options: (a) 20 N
(b) 50 N
(c) 100 N
(d) 2 N
Answer: Answer: (d) 2N
Solution: 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
Topic: 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)
Options: (a) 0.02
(b) 0.03
(c) 0.06
(d) 0.01
Answer: Answer: (c) 0.06
Solution: 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
Topic: 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.
Options: (a)kA/2
(b) kA
(c) μs mg
(d) zero
Answer: Answer: (a) fm=k A / 2
Solution: 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
Topic: 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.
Options: 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: Answer: (c) Statement-1 is true, statement-2 is false
Topic: 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 )
Options: (a) 2.0
(b) 4.0
(c) 1.6
(d) 2.5
Answer: Answer: (a) 2
Solution: Solution: μ=0.8
f=10 N
mg sin 30=10
mx 10×(1 / 2)=10
m=2 kg
Topic: 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-
Options: (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: Answer: (a) μ k=1-1 / n²
Solution: 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
Topic: 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
Options: (a) 2 sin Φ
(b) 2 cos Φ
(c) 2 tan Φ
(d) tan Φ
Answer: Answer: (c) 2 tan Φ
Solution: 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
Topic: 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)
Options: (a) 800 m
(b) 1000 m
(c) 100 m
(d) 400 m
Answer: Answer: (b) 1000
Solution: 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