164. A pulley whose string length is 2 meters and ball mass is 2kg, starting from a position where the string makes a $53^\circ$ angle with the vertical, is released from rest. The tension in the string at the moment it makes a $37^\circ$ angle with the vertical is how many Newtons? ($\sin 37^\circ = 0.6$, air resistance is negligible, $g = 10\,\frac{m}{s^2}$)
(1) $16$ (2) $20$ (3) $24$ (4) $36$

52 -- A car with mass $2\,\text{t}$ moves on a horizontal surface at a constant speed of $18\,\dfrac{\text{km}}{\text{h}}$ along a circular path with radius $20\,\text{m}$. How many Newtons is the centripetal force and which force provides it?

• [(1)] $2500$ -- kinetic friction force
• [(2)] $2500$ -- static friction force
• [(3)] $1250$ -- kinetic friction force
• [(4)] $1250$ -- static friction force

28. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m . The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N . The maximum possible value of angular velocity of ball (in radian/s) is [Figure]
(A) 9
(B) 18
(C) 27
(D) 36

ANSWER: D

1. A meter bridge is set-up as shown, to determine an unknown resistance ' $X$ ' using a standard 10 ohm resistor. The galvanometer shows null point when tapping-key is at 52 cm mark. The end-corrections are 1 cm and 2 cm respectively for the ends A and B . The determined value of ' $X$ ' is [Figure]
   (A) 10.2 ohm
   (B) 10.6 ohm
   (C) 10.8 ohm
   (D) 11.1 ohm

ANSWER:B

1. A $2 \mu \mathrm {~F}$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position 2 is [Figure]
   (A) $0 \%$
   (B) $20 \%$
   (C) $75 \%$
   (D) $80 \%$

ANSWER: D

SECTION - II (Total Marks : 16)

(Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE may be correct.

Initial angular velocity of a circular disc of mass $M$ is $\omega_1$. Then two small spheres of mass $m$ are attached gently to diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
(1) $\left( \frac{M+m}{M} \right) \omega_1$
(2) $\left( \frac{M+m}{m} \right) \omega_1$
(3) $\left( \frac{M}{M+4m} \right) \omega_1$
(4) $\left( \frac{M}{M+2m} \right) \omega_1$

A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
(1) $g x$
(2) $\frac { g R } { R - x }$
(3) $\frac { g ^ { 2 } } { R + x }$
(4) $\left( \frac { g R ^ { 2 } } { R + x } \right) ^ { 1 / 2 }$

The time period of an earth satellite in circular orbit is independent of
(1) the mass of the satellite
(2) radius of its orbit
(3) both the mass and radius of the orbit
(4) neither the mass of the satellite nor the radius of its orbit.

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to
(1) $R ^ { \left( \frac { n + 1 } { 2 } \right) }$
(2) $R ^ { \left( \frac { n - 1 } { 2 } \right) }$
(3) $R ^ { n }$
(4) $\mathrm { R } ^ { \left( \frac { \mathrm { n } - 2 } { 2 } \right) }$

An annular ring with inner and outer radii $R_1$ and $R_2$ is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, $F_1/F_2$ is
(1) $\frac{R_2}{R_1}$
(2) $\left(\frac{R_1}{R_2}\right)^2$
(3) 1
(4) $\frac{R_1}{R_2}$

A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length $s = t ^ { 3 } + 5$, where $s$ is in metres and $t$ is in seconds. The radius of the path is 20 m. The acceleration of 'P' when $t = 2 \mathrm {~s}$ is nearly
(1) $13 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(2) $12 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(3) $7.2 \mathrm {~m} / \mathrm { s } ^ { 2 }$
(4) $14 \mathrm {~m} / \mathrm { s } ^ { 2 }$

For a particle in uniform circular motion the acceleration $\vec { a }$ at a point $P ( R , \theta )$ on the circle of radius R is (here $\theta$ is measured from the $x$-axis)
(1) $- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } + \frac { v ^ { 2 } } { R } \sin \theta \hat { j }$
(2) $- \frac { v ^ { 2 } } { R } \sin \theta \hat { i } + \frac { v ^ { 2 } } { R } \cos \theta \hat { j }$
(3) $- \frac { v ^ { 2 } } { R } \cos \theta \hat { i } - \frac { v ^ { 2 } } { R } \sin \theta \hat { j }$
(4) $\frac { v ^ { 2 } } { R } \hat { i } + \frac { v ^ { 2 } } { R } \hat { j }$

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is $12 \mathrm { rad } \mathrm { s } ^ { - 1 }$, the magnitude of its angular momentum about a point on the ground right under the center of the circle is:
(1) $14.4 \mathrm {~kg} \mathrm {~m} ^ { 2 } \mathrm {~s} ^ { - 1 }$
(2) $11.52 \mathrm {~kg} \mathrm {~m} ^ { 2 } \mathrm {~s} ^ { - 1 }$
(3) $20.16 \mathrm {~kg} \mathrm {~m} ^ { 2 } \mathrm {~s} ^ { - 1 }$
(4) $8.64 \mathrm {~kg} \mathrm {~m} ^ { 2 } \mathrm {~s} ^ { - 1 }$

A conical pendulum of length $l$ makes an angle $\theta = 45 ^ { \circ }$ with respect to $Z$-axis and moves in a circle in the $X Y$ plane. The radius of the circle is 0.4 m and its center is vertically below $O$. The speed of the pendulum, in its circular path, will be - (Take $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$)
(1) $0.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) $0.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(3) $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distance of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $\left( \mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$
(1) 0.5
(2) 0.7
(3) 0.3
(4) 0.6

A particle is moving along a circular path with a constant speed of $10 \mathrm {~ms} ^ { - 1 }$. What is the magnitude of the change in velocity of the particle, when it moves through an angle of $60 ^ { \circ }$ around the centre of the circle?
(1) $10 \sqrt { 3 } \mathrm {~m} / \mathrm { s }$
(2) zero
(3) $10 \sqrt { 2 } \mathrm {~m} / \mathrm { s }$
(4) $10 \mathrm {~m} / \mathrm { s }$

A smooth wire of length $2 \pi r$ is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed $\omega$ about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of $\omega ^ { 2 }$ is equal to:
(1) $2 \mathrm {~g} / \mathrm { r }$
(2) $\frac { \sqrt { 3 } \mathrm {~g} } { 2 \mathrm { r } }$
(3) $2 g / ( r \sqrt { 3 } )$
(4) $( \mathrm { g } \sqrt { 3 } ) / \mathrm { r }$

A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? [Given: Mass of planet $= 8 \times 10 ^ { 22 } \mathrm {~kg}$, Radius of planet $= 2 \times 10 ^ { 6 } \mathrm {~m}$, Gravitational constant $\mathrm { G } = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } / \mathrm { kg } ^ { 2 }$ ]
(1) 17
(2) 9
(3) 13
(4) 11

A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $\mathrm { R } _ { \mathrm { e } }$. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it become $\sqrt { \frac { 3 } { 2 } }$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $R$. Value of $R$ is:
(1) $4 \mathrm { R } _ { \mathrm { e } }$
(2) $2.5 \mathrm { R } _ { \mathrm { e } }$
(3) $3 R _ { e }$
(4) $2 \mathrm { R } _ { \mathrm { e } }$

A body A of mass $m$ is moving in a circular orbit of radius $R$ about a planet. Another body B of mass $\frac { m } { 2 }$ collides with A with a velocity which is half $\left( \frac { \vec { v } } { 2 } \right)$ the instantaneous velocity $\vec { v }$ of A. The collision is completely inelastic. Then, the combined body:
(1) continues to move in a circular orbit
(2) Escapes from the Planet's Gravitational field
(3) Falls vertically downwards towards the planet
(4) starts moving in an elliptical orbit around the planet

A body of mass $\mathrm { m } = 10 \mathrm {~kg}$ is attached to one end of a wire of length 0.3 m . What is the maximum angular speed (in $\mathrm { rad } \mathrm { s } ^ { - 1 }$ ) with which it can be rotated about its other end in a space station without breaking the wire? [Breaking stress of wire $( \sigma ) = 4.8 \times 10 ^ { 7 } \mathrm {~N} \mathrm {~m} ^ { - 2 }$ and area of cross-section of the wire $= 10 ^ { - 2 } \mathrm {~cm} ^ { 2 }$ ]

A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with an angular speed $\omega$. Two particles having mass $m$ each are now attached at diametrically opposite points. The angular speed of the ring will become:
(1) $\omega \frac { M } { M + m }$
(2) $\omega \frac { M + 2 m } { M }$
(3) $\omega \frac { M } { M + 2 m }$
(4) $\omega \frac { M - 2 m } { M + 2 m }$

A particle of mass $m$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{\sqrt{2}}$. The speed of particle will be:
(1) $\sqrt{rg}$
(2) $\sqrt{2rg}$
(3) $\sqrt{\frac{rg}{2}}$
(4) $2\sqrt{rg}$

The time period of a satellite in a circular orbit of the radius $R$ is $T$. The period of another satellite in a circular orbit of the radius $9R$ is:
(1) $9T$
(2) $27T$
(3) $12T$
(4) $3T$

Two identical particles of mass 1 kg each go round a circle of radius $R$, under the action of their mutual gravitational attraction. The angular speed of each particle is:
(1) $\sqrt { \frac { G } { 2 R ^ { 3 } } }$
(2) $\frac { 1 } { 2 } \sqrt { \frac { G } { R ^ { 3 } } }$
(3) $\frac { 1 } { 2 R } \sqrt { \frac { 1 } { G } }$
(4) $\sqrt { \frac { 2 G } { R ^ { 3 } } }$

Four particles each of mass $M$, move along a circle of radius $R$ under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:
(1) $\frac { 1 } { 2 } \sqrt { \frac { G M } { R } (2 \sqrt { 2 } + 1)}$
(2) $\frac { 1 } { 2 } \sqrt { \frac { G M } { R ( 2 \sqrt { 2 } + 1 ) } }$
(3) $\frac { 1 } { 2 } \sqrt { \frac { G M } { R } (2 \sqrt { 2 } - 1)}$
(4) $\sqrt { \frac { G M } { R } }$

The centre of a wheel rolling on a plane surface moves with a speed $v _ { 0 }$. A particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt { x } v _ { 0 }$. Then the value of $x$ is $\_\_\_\_$ .