Chapter 10

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is $$ I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be $$ I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right) $$ When \(L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}\)

Two spheres of radius \(r\) and \(2 r\) are touching each other. The force of attraction between them is proportional to (a) \(r^{6}\) (b) \(r^{4}\) (c) \(r^{2}\) (d) \(r^{-2}\)

If the diameter of mars is \(6760 \mathrm{~km}\) and mass one-tenth that of earth. The diameter of earth is \(12742 \mathrm{~km}\). If acceleration due to gravity on earth is \(9.8 \mathrm{~ms}^{-2}\), the acceleration due to gravity on mars is (a) \(34.8 \mathrm{~ms}^{-2}\) (b) \(2.84 \mathrm{~ms}^{-2}\) (c) \(3.48 \mathrm{~ms}^{-2}\) (d) \(28.4 \mathrm{~ms}^{-2}\)

\(\mathrm{As}, I_{s}=\frac{2}{5} M R_{s}^{2}, I_{h}=\frac{2}{3} M R_{h}^{2}\) As \(\quad I_{s}=I_{h}\) \(\therefore \quad \frac{2}{5} M R_{s}^{2}=\frac{2}{3} M R_{h}^{2}\) \(\therefore\) \(\frac{R_{s}}{R_{h}}=\frac{\sqrt{5}}{\sqrt{3}}\)

A uniform ring of mass \(M\) and radius \(r\) is placed directly above a uniform sphere of mass \(8 M\) and of same radius \(R\). The centre of the ring is at a distance of \(d=\sqrt{3} R\) from the centre of the sphere. The gravitational attraction between the sphere and the ring is (a) \(\frac{G M^{2}}{R^{2}}\) (b) \(\frac{3 G M^{2}}{2 R^{2}}\) (c) \(\frac{2 G M^{2}}{\sqrt{2} R^{2}}\) (d) \(\frac{\sqrt{3} G M^{2}}{R^{2}}\)

As, \(\omega_{2}=\omega_{1}+\alpha t\) \(\therefore \quad 40 \pi=20 \pi+\alpha \times 10\) or \(\alpha=2 \pi \mathrm{rad} \mathrm{s}^{-2}\) \(\begin{aligned}&\text { From, } \quad \omega_{2}^{2}-\omega_{1}^{2} & =2 \alpha \theta \\\& & (40 \pi)^{2}-(20 \pi)^{2} & =2 \times 2 \pi \theta \\\&\Rightarrow \quad \theta & =\frac{1200 \pi^{2}}{4 \pi}=300 \pi\end{aligned}\) Number of rotations completed \(=\frac{\theta}{2 \pi}=\frac{300}{2 \pi}=150\)

Imagine a light planet revolving around a very massive star in a circular orbit of radius \(r\) with a period of revolution \(T\). If the gravitational force of attraction between the planet and the star is proportional to \(R^{-3 / 2}\), then \(T_{2}\) is proportional to (a) \(R^{3}\) (b) \(R^{5 / 2}\) (c) \(R^{3 / 2}\) (d) \(R^{7 / 2}\)

Linear acceleration for rolling, \(a=\frac{g \sin \theta}{\sqrt{1+\frac{K^{2}}{R^{2}}}}\) For cylinder, \(\frac{K^{2}}{R^{2}}=\frac{1}{2}\) \(\therefore \quad a_{\text {cylinder }}=\frac{2}{3} g \sin \theta\) For rotation, the torque $$ f R=l \alpha \cdot\left(M R^{2} \alpha\right) / 2 $$ (where, \(f=\) force of friction) But \(R \alpha=a \quad \therefore f=\frac{M}{2} a\) \(\therefore \quad f=\frac{M}{2} \cdot \frac{2}{3} g \sin \theta=\frac{M}{3} g \sin \theta\) \(\mu_{s}=f / N\), where \(N\) is normal reaction, \(\therefore\) $$ \mu_{s}=\frac{\frac{M}{3} g \sin \theta}{M g \cos \theta}=\frac{\tan \theta}{3} $$ \(\therefore\) For rolling without slipping of a roller down the inclined plane, \(\tan \theta \leq 3 \mu_{s}\).

Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon \(\quad\) [NCERT Exemplar] (a) will be elliptical (b) will not be strictly elliptical because the total gravitational force on it is not central (c) is not elliptical but will necessarily be a closed curve (d) deviates considerably from being elliptical due to influence of planets other than earth

The work that must be done in lifting a body of weight \(P\) from the surface of the earth to a height \(h\) is (a) \(\frac{P R h}{R-h}\) (b) \(\frac{R+h}{P R h}\) (c) \(\frac{P R h}{R+h}\) (d) \(\frac{R-h}{P R h}\)

Different points in earth are at slightly different distances from the sun and hence experience different forces due to gravitation. For a rigid body, we know that if various forces act at various points in it, the resultant motion is as if a net force acts on the CM (centre of mass) causing translation and a net torque at the CM causing rotation around an axis through the CM For the earth-sun system (approximating the earth as a uniform density sphere) \(\quad\) [NCERT Exemplar] (a) the torque is zero (b) the torque causes the earth to spin (c) the rigid body result is not applicable since the earth is not even approximately a rigid body (d) the torque causes the earth to move around the sun

A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is (Radius of earth \(=6400 \mathrm{~km}\) and \(g=9.8 \mathrm{~ms}^{-2}\) ) (a) \(11.2 \mathrm{kms}^{-1}\) (b) \(8 \mathrm{kms}^{-1}\) (c) \(3.2 \mathrm{kms}^{-1}\) (d) \(1.5 \mathrm{kms}^{-1}\)

Two astronauts have deserted their space ships in a region of space far from the gravitational attraction of any other body. Each has a mass of \(100 \mathrm{~kg}\) and they are \(100 \mathrm{~m}\) apart. They are initially at rest relative to one another. How long will it be before the gravitational attraction brings them \(1 \mathrm{~cm}\) closer together? (a) \(2.52\) days (b) \(1.41\) days (c) \(0.70\) days (d) \(0.41\) days

If three particles each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the forces exerted by this system on another particle of mass \(M\) placed (i) at the mid point of a side and (ii) at the centre of the triangle are respectively (a) 0,0 (b) \(\frac{4 G M^{2}}{3 a^{2}}, 0\) (c) \(0, \frac{4 G M^{2}}{3 a^{2}}\) (d) \(\frac{3 G M^{2}}{a^{2}}, \frac{G M^{2}}{a^{2}}\)

The gravitational attraction between the two bodies increases when their masses are (a) reduced and distance is reduced (b) increased and distance is reduced (c) reduced and distance is increased (d) increased and distance is increased

A spherical hollow is made in a lead sphere of radius \(R\) such that its surface touches the outside surface of the lead sphere and passes through the centre. The mass of the lead sphere before hollowing was \(M\). The force of attraction that this sphere would exert on a particle of mass \(m\) which lies at a distance \(d(>R)\) from the centre of the lead sphere on the straight line joining the centres of the sphere and the hollow is (a) \(\frac{G M m}{d^{2}}\) (b) \(\frac{G M m}{8 d^{2}}\) (c) \(\frac{G M m}{d^{2}}\left[1+\frac{1}{8\left(1+\frac{R}{2 d}\right)}\right]\) (d) \(\frac{G M m}{d^{2}}\left[1-\frac{1}{8\left(1-\frac{R}{2 d}\right)^{2}}\right]\)

A star \(2.5\) times the mass of the sun and collasped to a size of \(12 \mathrm{~km}\) rotates with a speed of \(1.2 \mathrm{rev} / \mathrm{s}\). (Extremely compact stars of this kind are known as neutron stars. Certain observed stellar objects called pulsars are belived to belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (Mass of the sun \(=2 \times 10^{30} \mathrm{~kg}\) ). (a) Yes (b) No (c) Sometimes yes and sometimes no (d) Can not be said

A spherical hollow is made in a lead sphere of radius \(R\) such that its surface touches the outside surface of the lead sphere and passes through the centre. The mass of the lead sphere before hollowing was \(M\). The force of attraction that this sphere would exert on a particle of mass \(m\) which lies at a distance \(d(>R)\) from the centre of the lead sphere on the straight line joining the centres of the sphere and the hollow is (a) \(\frac{G M m}{d^{2}}\) (b) \(\frac{G M m}{8 d^{2}}\) (c) \(\frac{G M m}{d^{2}}\left[1+\frac{1}{8\left(1+\frac{R}{2 d}\right)}\right]\) (d) \(\frac{G M m}{d^{2}}\left[1-\frac{1}{8\left(1-\frac{R}{2 d}\right)^{2}}\right]\)

According to the theorem of parallel axes, moment of inertia of disc about an axis passing through \(K\) and perpendicular to plane of \(\operatorname{disc}_{,}\) \(=\frac{1}{2} M R^{2}+M R^{2}=\frac{3}{2} M R^{2}\) Total moment of inertia of the system \(=\frac{3}{2} M R^{2}+m(2 R)^{2}+m(\sqrt{2} R)^{2}+m(\sqrt{2} R)^{2}\) \(=3(M+16 m) \frac{R^{2}}{2}\)

If suppose moon is suddenly stopped and then released (given radius of moon is one-fourth the radius of earth) and the acceleration of moon with respect to earth is \(0.0027 \mathrm{~ms}^{-2}\) ), then the acceleration of the moon just before striking the earth's surface is (Take \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(0.0027 \mathrm{~ms}^{-2}\) (b) \(5.0 \mathrm{~ms}^{-2}\) (c) \(6.4 \mathrm{~ms}^{-2}\) (d) \(10 \mathrm{~ms}^{-2}\)

Distance between the centres of two stars is \(10 a\). The masses of these stars are \(M\) and \(16 M\) and their radii \(a\) and \(2 a\) respectively. A body of mass \(m\) is fired straight from the surface of the larger star towards the smaller star. The minimum initial speed for the body to reach the surface of smaller star is (a) \(\frac{2}{3} \sqrt{\frac{G M}{a}}\) (b) \(\frac{3}{2} \sqrt{\frac{5 G M}{a}}\) (c) \(\frac{2}{3} \sqrt{\frac{5 G M}{a}}\) (d) \(\frac{3}{2} \sqrt{\frac{G M}{a}}\)

Three particles each of mass \(m\) rotate in a circle of radius \(r\) with uniform angular speed \(\omega\) under their mutual gravitational attraction. If at any instant the points are on the vertex of an equilateral of side \(L\), then angular velocity \(\omega\) is (a) \(\sqrt{\frac{2 G m}{L^{3}}}\) (b) \(\sqrt{\frac{3 G m}{L^{3}}}\) (c) \(\sqrt{\frac{5 G m}{L^{3}}}\) (d) \(\sqrt{\frac{G m}{L^{3}}}\)

As, \(L=r P\) \(\Rightarrow \quad \quad \log _{e} L=\log _{e} P+\log _{e} r\) If graph is drawn between \(\log _{e} L\) and \(\log _{e} P\) then, it will be straight line which will not pass through the origin because of presence of constant in the equation.

A thief stole a box full of valuable articles of weight \(w\) and while carrying it on his head jumped down from a wall of height \(h\) from the ground. Before he reaches the ground, he experienced a load (a) zero (b) \(w / 2\) (c) \(w\) (d) \(2 w\)

Assuming the earth to be a sphere of uniform mass density, how much would body weigh half way down to the centre of earth if it weighed \(250 \mathrm{~N}\) on the surface? (a) \(225 \mathrm{~N}\) (b) \(325 \mathrm{~N}\) (c) \(100 \mathrm{~N}\) (d) \(125 \mathrm{~N}\)

Two satellites \(S_{1}\) and \(S_{2}\) revolve around a planet in coplanar circular orbits in the same sense. Their periods of revolution are \(1 \mathrm{~h}\) and \(8 \mathrm{~h}\) respectively. The radius of orbit of \(S_{1}\) is \(10^{4} \mathrm{~km}\). When \(S_{2}\) is closest to \(S_{1}\), the speed of \(S_{2}\) relative to \(S_{1}\) is (a) \(\pi \times 10^{4} \mathrm{kmh}^{-1}\) (b) \(2 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (c) \(3 \pi \times 10^{4} \mathrm{kmh}^{-1}\) (d) \(4 \pi \times 10^{4} \mathrm{kmh}^{-1}\)

Let \(L_{1}, L_{2}\) and \(r_{1}, r_{2}\) are the angular momenta and position vectors of the particles at that instant about any arbitrary point \(O .\) Angular momentum of the particles, $$\mathbf{L}_{1}=\mathbf{r}_{1} \times m \mathbf{v} \text { and } \mathbf{L}_{2}=\mathbf{r}_{2} \times m \mathbf{v}$$ It resultant angular momentum of the system is \(\mathbf{L}\), then $$\mathbf{L}=\mathbf{L}_{1}+\mathbf{L}_{2}=\mathbf{r}_{1} \times m \mathbf{v}+\left(-\mathbf{r}_{2}+m \mathbf{v}\right)$$ Negative sign shown that both particles are moving in opposite directions. \(\begin{aligned} |\mathbf{L}| &=\left|\mathbf{L}_{1}\right|-\left|\mathbf{L}_{2}\right| \\ &=m v r_{1} \sin \theta_{1}-m v r_{2} \sin \theta_{2} \\ &=m v\left(r_{1} \sin \theta_{1}-r_{2} \sin \theta_{2}\right) \end{aligned}\) where \(\theta_{1}\) and \(\theta_{2}\) are the angles between \(\mathbf{r}_{1}, \mathbf{v}\) and \(\mathbf{r}_{2}, \mathbf{v}\) respectively. When particles changes their position with time, their direction of motion (v) remains unchanged and therefore distances \(O M=r_{1} \sin \theta_{1}\) and \(O N=r_{2} \sin \theta_{2}\) remains same. But \(\quad O M-O N=M N=d \quad\) (Given) \(\therefore \quad r_{1} \sin \theta_{1}-r_{2} \sin \theta_{2}=d\) (ii) From Eqs. (i) and (ii), we get \(|\mathbf{L}|=m v d\) It is constant with time. The direction of \(\mathbf{L}\) is perpendicular to the plane of \(\mathbf{r}\) and \(\mathbf{v}\) and is inward to the plane of paper, which also remains unchanged with time.

The maximum vertical distance through which a full dressed astronaut can jump on the earth is \(0.5 \mathrm{~m}\). Estimate the maximum vertical distance through which he can jump on the moon, which has a mean density \(2 / 3 \mathrm{rd}\) that of earth and radius one quarter that of the earth (a) \(1.5 \mathrm{~m}\) (b) \(3 \mathrm{~m}\) (c) \(6 \mathrm{~m}\) (d) \(7.5 \mathrm{~m}\)

A straight rod of length \(L\) extends from \(x=a\) to \(x=L+a\). Find the gravitational force it, exerts on a point mass \(m\) at \(x=0\) if the linear density of rod \(\mu=A+B x^{2}\) (a) \(\mathrm{Gm}\left[\frac{A}{\mathrm{a}}+\mathrm{BL}\right]\) (b) \(G m\left[A\left(\frac{1}{a}-\frac{1}{a+L}\right)+B L\right]\) (c) \(G m\left[B L+\frac{A}{a+L}\right]\) (d) \(G m\left[B L-\frac{A}{a}\right]\)

The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity \(\quad\) [NCERT Exemplar] (a) will be directed towards the centre but not the same everywhere (b) will have the same value everywhere but not directed towards the centre (c) will be same everywhere in magnitude directed towards the centre (d) cannot be zero at any point

An object weighs \(10 \mathrm{~N}\) at the north-pole of the earth. In a geostationary satellite distant \(7 R\) from the centre of earth (of radius \(R\) ), what will be its true weight? (a) \(3 \mathrm{~N}\) (b) \(5 \mathrm{~N}\) (c) \(2 \mathrm{~N}\) (d) \(0.2 \mathrm{~N}\)

The masses and radii of the earth and moon are \(M_{1}, R_{1}\) and \(M_{2}, R_{2}\) respectively. Then centres are distance \(d\) apart. The minimum velocity with which a particle of mass \(M\) should be projected from a point midway between their centres so that it escapes to infinity is (a) \(2 \sqrt{\frac{G}{d}\left(M_{1}+M_{2}\right)}\) (b) \(2 \sqrt{\frac{2 G}{d}\left(M_{1}+M_{2}\right)}\) (c) \(2 \sqrt{\frac{G M}{d}\left(M_{1}+M_{2}\right)}\) (d) \(2 \sqrt{\frac{G M\left(M_{1}+M_{2}\right)}{d\left(R_{1}+R_{2}\right)}}\)

At a given place where, acceleration due to gravity is \(g \mathrm{~ms}^{-2}\), a sphere of lead of density \(d \mathrm{kgm}^{-3}\) is gently released in a column of liquid of density \(\rho \mathrm{kgm}^{-3}\). If \(d>\rho\), the sphere will (a) fall vertically with an acceleration of \(\mathrm{g} \mathrm{ms}^{-2}\) (b) fall vertically with no acceleration (c) fall vertically with an acceleration \(g\left(\frac{d-\rho}{d}\right)\) (d) fall vertically with an acceleration \(\rho / d\)

Two equal mases \(m\) and \(m\) are hung from balance whose scale pans differ in vertical height by \(h\). Calculate the error in weighing, if any, in terms of density of earth \(\rho\). (a) \(\frac{2}{3} \pi \rho R^{3} \mathrm{Gm}\) (b) \(\frac{8}{3} \pi \rho G m h\) (c) \(\frac{8}{3} \pi \rho R^{3} G m\) (d) \(\frac{4}{3} \pi \rho G m^{2} h\)

If satellite is revolving around a planet of mass \(M\) in an elliptical orbit of semi-major axis \(a\), find the orbital speed of the satellite when it is at a distance \(r\) from the focus. (a) \(v^{2}=G M\left[\frac{2}{r}-\frac{1}{a}\right]\) (b) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a}\right]\) (c) \(v^{2}=G M\left[\frac{2}{r^{2}}-\frac{1}{a^{2}}\right]\) (d) \(v^{2}=G\left[\frac{2}{r}-\frac{1}{a}\right]\)

There is a mine of depth about \(2.0 \mathrm{~km}\). In this mine the conditions as compared to those at the surface are (a) lower air pressure, higher acceleration due to gravity (b) higher air pressure, lower acceleration due to gravity (c) higher air pressure, higher acceleration due to gravity (d) lower air pressure, lower acceleration due to gravity

A clock \(S\) is based on oscillation of a spring and a clock \(P\) is based on pendulum motion. Both clock run at the same rate on earth. On a planet having the same density as earth but twice the radius, (a) \(S\) will run faster than \(P\) (b) \(P\) will run faster than \(S\) (c) both will run at the same rate as on the earth (d) both will run at the same rate which will be different from that on the earth

The potential energy of gravitational interaction of a point mass \(m\) and a thin uniform rod of mass \(M\) and length \(l\), if they are located along a straight line at distance a from each other is (a) \(U=\frac{G M m}{a} \log _{e}\left(\frac{a+l}{a}\right)\) (b) \(U=G M m\left(\frac{1}{a}-\frac{1}{a+l}\right)\) (c) \(U=\frac{G M m}{l} \log _{e}\left(\frac{a+l}{a}\right)\) (d) \(U=-\frac{G M m}{a}\)

If the radius of the earth were to shrink by \(1 \%\) its mass remaining same, the acceleration due to gravity on the earth's surface would (a) decrease by \(2 \%\) (b) remain unchanged (c) increase by \(2 \%\) (d) become zero

\(X_{1}+x_{2}=r\) \(\ldots\)..(i) and \(m_{1} x_{1}=m_{2} x_{2}\) ...(ii) From Eqs. (i) and (ii), \(x_{1}=\frac{m_{2} r}{m_{1}+m_{2}}\) and \(x_{2}=\frac{m_{1} r}{m_{1}+m_{2}}\) \(\therefore\) \(I_{A B}=m_{1} x_{1}^{2}+m_{2} x_{2}^{2}=\frac{m_{1} m_{2} r^{2}}{m_{1}+m_{2}}\)

Two spherical planets \(A\) and \(B\) have same mass but densities in the ratio \(8: 1\). For these planets, the ratio of acceleration due to gravity at the surface of \(A\) to its value at the surface of \(B\) is (a) \(1: 4\) (b) \(1: 2\) (c) \(4: 1\) (d) \(8: 1\)

How much energy will be necessary for making a body of \(500 \mathrm{~kg}\) escape from the earth? \(\left(g=9.8 \mathrm{~ms}^{-2}\right.\), radius of earth \(\left.=6.4 \times 10^{6} \mathrm{~m}\right)\) (a) About \(9.8 \times 10^{6} \mathrm{~J}\) (b) About \(6.4 \times 10^{8} \mathrm{~J}\) (c) About \(3.1 \times 10^{10} \mathrm{~J}\) (d) About \(27.4 \times 10^{12} \mathrm{~J}\)

As there is no external torque, angular momentum will remain constant. When the tortoise moves from \(A\) to \(C\), figure, moment of inertia of the platform and tortoise decreases. Therefore, angular velocity of the system increases. When the tortoise moves from \(C\) to \(B\), moment of inertia increases. Therefore, angular velocity decreases. If, \(M=\) mass of platform \(R=\) radius of platform \(m=\) mass of tortoise moving along the chord \(A B\) \(\mathrm{a}=\) perpendicular distance of \(O\) from \(A B\). Initial angular momentum, \(l_{1}=m R^{2}+\frac{M R^{2}}{2}\) At any time \(t\), let the tortoise reach \(D\) moving with velocity \(\underline{v}\). \(\therefore\) \(A D=v t\) As $$\begin{aligned} &A C=\sqrt{R^{2}-a^{2}} \\ &D C=A C-A D=\left(\sqrt{R^{2}-a^{2}}-v t\right) \end{aligned}$$ \(\therefore\) $$O D=r=a^{2}+\left[\sqrt{R^{2}-a^{2}}-v t\right]^{2}$$ Angular momentum at time \(t\) $$ I_{2}=m r^{2}+\frac{M R^{2}}{2} $$ As angular momentum is conserved \(\therefore\) $$l_{1} \omega_{0}=I_{2} \omega(t)$$ This shows that variation of \(\omega(t)\) with time is non-linear.

The height at which the acceleration due to gravity decreases by \(36 \%\) of its value on the surface of the earth. (The radius of the earth is \(R\) ). (a) \(\frac{R}{6}\) (b) \(\frac{R}{4}\) (c) \(\frac{R}{2}\) (d) \(\frac{2}{3} R\)

Two identical satellites are at \(R\) and \(7 R\) away from earth surface, the wrong statement is ( \(R=\) Radius of earth) (a) ratio of total energy will be \(y\) (b) ratio of kinetic energies will be \(y\) (c) ratio of potential energies will be \(y\) (d) ratio of total energy will be 4 but ratio of potential and kinetic energy will be 1

If the value of \(g\) acceleration due to gravity at earth surface is \(10 \mathrm{~ms}^{-2}\), its value in \(\mathrm{ms}^{-2}\) at the centre of the earth, which is assumed to be a sphere of radius \(R\) metre and uniform mass density is (a) 5 (b) \(10 / R\) (c) \(10 / 2 R\) (d) zero

A particle is projected vertically upwards from the surface of earth (radius \(R_{e}\) ) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of earth is (a) \(R_{e}\) (b) \(2 R_{e}\) (c) \(3 R_{e}\) (d) \(4 R_{e}\)

Moment of inertia of the cylinder about an axis perpendicular to the axis of the cylinder and passing through the centre is $$ I=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ If \(\rho\) is volume density of the cylinder, then $$ M=\left(\pi R^{2} L\right) \rho=\text { constant } $$ \(\therefore\) $$ L=\frac{M}{\pi R^{2} \rho} $$ Put in Eq. (i) $$ I=M\left(\frac{R^{2}}{4}+\frac{M^{2}}{12 \pi^{2} R^{4} \rho^{4}}\right) $$ For \(/\) to be minimum, \(\frac{d l}{d R}=0\) $$ \begin{aligned} &\frac{d l}{d R}=M\left(\frac{R}{2}-\frac{M^{2}}{3 \pi^{2} R^{5}}\right)=0 \\ &\frac{R}{2}=\frac{M^{2}}{3 \pi^{2} \rho^{2} R^{5}} \quad \text { or } \quad R^{6}=\frac{2 M^{2}}{3 \pi^{2} \rho^{2}} \end{aligned} $$ Using Eq. (ii), \(\quad R^{6}=\frac{2 \pi^{2} R^{4} L^{2} \rho^{2}}{3 \pi^{2} \rho^{2}}\) Or \(R^{2}=\frac{2}{3} L^{2}\) or \(\frac{L^{2}}{R^{2}}=\frac{3}{2}\) Or \(\frac{L}{R}=\sqrt{3 / 2}\)

A point \(P(R \sqrt{3}, 0,0)\) lies on the axis of a ring of mass \(M\) and radius \(R\). The ring is located in \(y-z\) plane with its centre at origin \(O .\) A small particle of mass \(m\) starts from \(P\) and reaches \(O\) under gravitational attraction only. Its speed at \(O\) will be (a) \(\sqrt{\frac{G M}{R}}\) (b) \(\sqrt{\frac{G m}{R}}\) (c) \(\sqrt{\frac{G M}{\sqrt{2} R}}\) (d) \(\sqrt{\frac{G m}{\sqrt{2} R}}\)

If \(M\) mass of the square plate before cutting the holes, then mass of portion of each hole. $$ m=\frac{M}{16 R^{2}} \times \pi R^{2}=\frac{\pi}{16} M $$ \(\therefore\) Moment of inertia of remaining portion $$ \begin{aligned} I &=I_{\text {square }}-4 l_{\text {hole }} \\ &=\frac{M}{12}\left(16 R^{2}+16 R^{2}\right)-4\left[\frac{m R^{2}}{2}+m(\sqrt{2} R)^{2}\right] \\ &=\frac{M}{12} \times 32 R^{2}-10 \mathrm{mR}^{2} \\ &=\frac{8}{3} M R^{2}-\frac{10 \pi}{16} M R^{2}=\left(\frac{8}{3}-\frac{10 \pi}{16}\right) M R^{2} \end{aligned} $$