Chapter 10

A spaceship is stationed on mars. How much energy must be expended on the spaceship to rocket it out of the solar system? Mass of the spaceship \(=1000 \mathrm{~kg}\); mass of sun \(=2 \times 10^{30} \mathrm{~kg} ;\) mass of mars \(=6.4 \times 10^{23} \mathrm{~kg} ;\) radius of mars \(=3395 \mathrm{~km}\); radius of orbit of mars \(=228 \times 10^{8} \quad \mathrm{~km}, \quad G=6.67 \times 10^{-11}\) \(\mathrm{N}-\mathrm{m}^{2} / \mathrm{kg}^{2}\). (a) \(3 \times 10^{11} \mathrm{~J}\) (b) \(4 \times 10^{11} \mathrm{~J}\) (c) \(33 \times 10^{11} \mathrm{~J}\) (d) None of these

\(320 \mathrm{~km}\) above the surface of earth, the value of acceleration due to gravity is nearly \(90 \%\) of its value on the surface of the earth. Its value will be \(95 \%\) of the value on the earth's surface (a) nearly \(160 \mathrm{~km}\) below the earth's surface (b) nearly \(80 \mathrm{~km}\) below the earth's surface (c) nearly \(640 \mathrm{~km}\) below the earth's surface (d) nearly \(320 \mathrm{~km}\) below the earth's surface

A planet of mass \(m\) is revolving round the sun (of mass \(m_{s}\) ) in an elliptical orbit. If \(\mathbf{v}\) is the velocity of the planet when its position vector from sun \(r\) then if the planet rotates in counter clockwise direction then areal velocity has direction (a) given by "Right Hand Thumb Rule" (b) given by "Left Hand Thumb Rule" (c) normal to the plane of orbit upwrads (d) normal to the plane of orbit downwards

The acceleration due to gravity at a height \(1 / 20\) th of the radius of the earth above the earth surface is \(9 \mathrm{~ms}^{-2}\). Its value at a point at an equal distance below the surface of the earth in \(\mathrm{ms}^{-2}\) is about (a) \(8.5\) (b) \(9.5\) (c) \(9.8\) (d) \(11.5\)

If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube law (a) planets will not have elliptic orbits (b) circular orbits of planets is not possible (c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic (d) there will be no gravitational force inside a spherical shell of uniform density

At a distance \(320 \mathrm{~km}\) above the surface of earth, the value of acceleration due to gravity will be lower than its value on the surface of the earth by nearly (radius of earth \(=6400 \mathrm{~km}\) ) (a) \(2 \%\) (b) \(6 \%\) (c) \(10 \%\) (d) 1496

There have been suggestions that the value of the gravitational constant \(G\) becomes smaller when considered over very large time period (in billions of years) in the future. If that happens, for our earth, (a) nothing will change (b) we will become hotter after billions of years (c) we will be going around but not strictly in closed orbits (d) after sufficiently long time we will leave the solar system

The depth from the surface of the earth of radius \(R\) at which the acceleration due to gravity will be \(75 \%\) of the value on the surface of the earth is (a) \(R / 4\) (b) \(R / 2\) (c) \(3 \mathrm{R} / 4\) (d) \(R / 8\)

Supposing Newton's law of gravitation for gravitation forces \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) between two masses \(m_{1}\) and \(m_{2}\) at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) read $$ \mathbf{F}_{1}=-\mathbf{F}_{2}=-\frac{\mathbf{r}_{12}}{r_{12}^{3}} G M_{0}^{2}\left(\frac{m_{1} m_{2}}{M_{0}^{2}}\right)^{n} $$ where \(M_{0}\) is a constant of dimension of mass, \(\mathbf{r}_{12}=\mathbf{r}_{1}-\mathbf{r}_{2}\) and \(n\) is a number. In such \(a\) case (a) the acceleration due to gravity on earth will be different for different objects (b) none of the three laws of Kepler will be valid (c) only the third law will become invalid (d) for \(n\) negative, an object lighter than water will sink in water

A body is imparted a velocity \(v\) from surface of the earth. If \(v_{0}\) is orbital velocity and \(v_{e}\) be the escape velocity, then for (a) \(v=v_{0}\), the body follows a circular track around the earth (b) \(v>v_{0}\), but \(v_{e}\), the body follows hyperbolic path and escapes the gravitational pull of the earth

Mass of moon is \(7.34 \times 10^{22} \mathrm{~kg}\). If the acceleration due to gravity on the moon is \(1.4 \mathrm{~ms}^{-2}\), the radius of the moon is \(\left(G=6.667 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right)\) (a) \(0.56 \times 10^{4} \mathrm{~m}\) (b) \(1.87 \times 10^{6} \mathrm{~m}\) (c) \(1.92 \times 10^{6} \mathrm{~m}\) (d) \(1.01 \times 10^{8} \mathrm{~m}\)

If the mass of sun were ten times smaller and gravitational constant \(G\) were ten times larger in magnitudes (a) walking on ground would became more difficult (b) the acceleration due to gravity on earth will not change (c) raindrops will fall much faster (d) airplanes will have to travel much faster

The ratio of acceleration due to gravity at a height \(h\) above the surface of the earth and at a depth \(h\) below the surface of the earth for \(h<\) radius of earth (a) is constant (b) increases linearly with \(h\) (c) decreases linearly with \(h\) (d) decreases parabolically with \(h\)

Here, \(\begin{aligned} n_{0} &=15 \mathrm{rps} ; n=5 \mathrm{rps} \\ \theta &=50 \text { revolution }=50 \times 2 \text { rad } \end{aligned}\) From, \(\begin{aligned} \omega^{2}-\omega_{0}^{2} &=2 \alpha \theta \\ \alpha &=\frac{\omega^{2}-\omega_{0}^{2}}{2 \theta}=\frac{4 \pi^{2}\left(n^{2}-n_{0}^{2}\right)}{2 \theta} \\ \alpha &=\frac{4 \pi^{2}\left(5^{2}-15^{2}\right)}{2 \times 5 \times 2 \pi}=-4 \pi \mathrm{rad} \mathrm{s}^{-} \end{aligned}\)

At what height in \(\mathrm{km}\) over the earth's pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be \(6400 \mathrm{~km}\) ) (a) 32 (b) 64 (c) 80 (d) \(1.253\)

If a man weighs \(90 \mathrm{~kg}\) on the surface of earth, the height above the surface of the earth of radius \(R\), where the weight is \(30 \mathrm{~kg}\), is (a) \(0.73 R\) (b) \(R[\sqrt{3}\) (c) \(R / 3\) (d) \(\sqrt{3} R\)

Let \(M\) and \(R\) be the mass of radius of the solid sphere and hollow cylinder. Moment of inertia of the hollow cylinder about its axis of symmetry, $$ I_{1}=M R^{2} $$ Moment of linear of the solid sphere about its diameter $$ I_{2}=\frac{2}{5} M R^{2} $$ Let torque \(\tau\) of triangle magnitude be applied on hollow cylinder and solid sphere. the angular acceleration produced in it are \(\alpha_{1}\) and \(\alpha_{2}\) respectively. \(\therefore\) \(\tau=I_{1} \alpha_{1}\) and \(\quad \tau=I_{2} \alpha_{2}\) Therefore, $$ \begin{gathered} I_{1} \alpha_{1}=I_{2} \alpha_{2} \\ \frac{\alpha_{1}}{\alpha_{2}}=\frac{l_{2}}{I_{1}}=\frac{\frac{2}{5} M R^{2}}{M R^{2}}=\frac{2}{5} \end{gathered} $$ or or $$ \alpha_{2}=\frac{5}{2} \alpha_{1} $$ $$ =2.5 \alpha_{1} $$ Let after time \(t, \omega_{1}\) and \(\omega_{2}\) be the angular speeds of the hollow cylinder and solid sphere respectively. \(\therefore\) $$ \omega_{1}=\omega_{0}+\alpha_{1} t $$ and $$ \begin{aligned} &\omega_{2}=\omega_{0}+\alpha_{2} t \\ &=\omega_{0}+2.5 \alpha_{1} t \end{aligned} $$ From Eqs. (ii) and (iii), we get $$ \omega_{2}>\omega_{1} $$ Therefore, solid sphere will acquire a greater angular speed after a given time.

Two equal masses \(m\) and \(m\) are hung from a balance whose scale pan differs in vertical height by \(h / 2\). 'The error in weighing in terms of density of the earth \(\rho\) is (a) \(\frac{1}{3} \pi G \rho m h\) (b) \(\pi G \pi m h\) (c) \(\frac{4}{3} \pi G \rho m h\) (d) \(\frac{8}{3} G \rho m h\)

The radius of the earth is \(6400 \mathrm{~km}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) in order that a body of \(5 \mathrm{~kg}\) weight zero at the equator the angular speed of the earth is (a) \(1 / 80 \mathrm{rad} / \mathrm{s}\) (b) \(1 / 400 \mathrm{rad} / \mathrm{s}\) (c) \(1 / 800 \mathrm{rad} / \mathrm{s}\) (d) \(1 / 1600 \mathrm{rad} / \mathrm{s}\)

What should be the angular speed of earth in \(\mathrm{rads}^{-1}\) so that a body of \(5 \mathrm{~kg}\), weighs zero at the equator? (Take \(g=10 \mathrm{~ms}^{-2}\) and radius of earth \(=6400 \mathrm{~km}\) ). (a) \(1 / 1600\) (b) \(1 / 800\) (c) \(1 / 400\) (d) \(1 / 80\)

When a person is high up on the ladder. Then a large torque is produced due to his weight about the point of contact between the ladder and the floor whereas when he starts climbing up the torque is small, Due to this reason the ladder is more opt to slip when one is high up on it.

At what height above the earth's surface, does the force of gravity decrease by \(10 \% ?\) The radius of the earth is \(6400 \mathrm{~km} ?\) (a) \(345.60 \mathrm{~km}\) (b) \(687.20 \mathrm{~km}\) (c) \(1031.8 \mathrm{~km}\) (d) \(12836.80 \mathrm{~km}\)

The value of \(g\) on the earth's surface is \(980 \mathrm{cms}^{-2}\). Its value at a height of \(64 \mathrm{~km}\) from the earth's surface is (Radius of the earth \(R=6400 \mathrm{~km}\) ) (a) \(960.40 \mathrm{cms}^{-2}\) (b) \(984.90 \mathrm{cms}^{-2}\) (c) \(982.45 \mathrm{cms}^{-2}\) (d) \(977.55 \mathrm{cms}^{-2}\)

Question No. 48 to 56 arc Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion Gravitational force between two particles is negligibly small compared to the electrical force. Reason The electrical force is experienced by charged particles only.

The speed of earth's rotation about its axis is \(\omega\). Its speed is increased to \(x\) times to make the effective acceleration due to gravity equal to zero at the equator, then value of \(x\) is around \(\left(g=10 \mathrm{~ms}^{-2} ; R=6400 \mathrm{~km}\right)\) (a) 1 (b) \(8.5\) (c) 17 (d) 34

Assertion There is no effect of rotation of earth on acceleration due to gravity at poles. Reason Rotation of earth is about polar axis.

For a body lying on the equator to appear weightless, what should be the angular speed of the earth? (Take \(g=10 \mathrm{~ms}^{-2}\); radius of earth \(=6400 \mathrm{~km}\) ) (a) \(0.125 \mathrm{rads}^{-1}\) (b) \(1.25 \mathrm{rads}^{-1}\) (c) \(1.25 \times 10^{-3} \mathrm{rads}^{-1}\) (d) \(1.25 \times 10^{-2} \mathrm{rads}^{-1}\)

Assertion When distance between two bodies is doubled and mass of each body is also doubled. Gravitational force between them remains the same. Reason According to Newton's law of gravitation, force is directly proportional to mass of bodies and inversely proportional to distance between them.

Moment of inertia of the square plate about an axis passing through the centre and perpendicular to its plane is $$ I=\frac{m\left(a^{2}+a^{2}\right)}{12}=\frac{m a^{2}}{6} $$

The gravitational field due to a mass distribution is \(I=k / x^{3}\) in the \(x\)-direction ( \(k\) is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance \(x / \sqrt{2}\) is (a) \(k / x\) (b) \(k / 2 x\) (c) \(k / x^{2}\) (d) \(k / 2 x^{2}\)

Angular accelerations is time derivative of angular speed and angular speed is time derivative of angular displacement. By definition \(\alpha=\frac{d \omega}{d t}\) i.e., $$ d \omega=\alpha d t $$ So, if in time \(t\) the angular speed of a body changes from \(\omega_{0}\) to \(\omega\) $$ \int_{\omega_{0}}^{\omega} d \omega=\int_{0}^{t} \alpha d t $$ If \(\alpha\) is constant $$ \omega-\omega_{0}=\alpha t $$ Now, as by definition $$ \omega=\frac{d \theta}{d t} $$ and $$ \theta=\frac{d \omega}{d t} $$ Eq. (i) becomes i.e., $$ \begin{aligned} &\frac{d \theta}{d t}=\omega_{0}+\alpha t \\ &d \theta=\left(\omega_{0}+a t\right) d t \end{aligned} $$ So, if in time \(t\) angular displacenent is \(\theta\) $$ \begin{gathered} \int_{0}^{\theta} d \theta=\int_{0}^{t}\left(\omega_{0}+\alpha t\right) d t \\\ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} \end{gathered} $$ Given, \(\alpha=3.0 \mathrm{rads}^{-2}, \omega_{0}=2.0 \mathrm{rads}^{-1}, t=2 \mathrm{~s}\) Hence, or \(\theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2}\) \(\theta=4+6=10 \mathrm{rad}\) Eqs. (i) and (ii) are similar to first and second equations of linear motion.So, if in time \(t\) angular displacenent is \(\theta\) $$ \begin{gathered} \int_{0}^{\theta} d \theta=\int_{0}^{t}\left(\omega_{0}+\alpha t\right) d t \\\ \theta=\omega_{0} t+\frac{1}{2} \alpha t^{2} \\ \text { Given, } \alpha=3.0 \mathrm{rads}^{-2}, \omega_{0}=2.0 \mathrm{rads}^{-1}, t=2 \mathrm{~s} \end{gathered} $$ Hence, $$ \theta=2 \times 2+\frac{1}{2} \times 3 \times(2)^{2} $$ or \(\theta=4+6=10 \mathrm{rad}\) Eqs. (i) and (ii) are similar to first and second equations of linear mot?on.

Assertion If earth suddenly stops rotating about its axis then the value of acceleration due to gravity will become same at all the places. Reason The value of acceleration due to gravity is independent of rotation of earth.

Torque due to central force is zero As \(\quad \tau=\frac{d}{d t}(L)=0\) \(\therefore\) \(L=\) constant

The distance between the earth and the moon is \(3.85 \times 10^{\mathrm{B}} \mathrm{m}\). At what distance from the earth's centre, the intensity of gravitational field will be zero? The masses of earth and moon are \(5.98 \times 10^{24} \mathrm{~kg}\) and \(7.35 \times 10^{22} \mathrm{~kg}\) respectively. (a) \(3.47 \times 10^{8} \mathrm{~m}\) (b) \(0.39 \times 10^{8} \mathrm{~m}\) (c) \(1.82 \times 10^{8} \mathrm{~m}\) (d) None of these

Two bodies of masses \(100 \mathrm{~kg}\) and \(1000 \mathrm{~kg}\) are separated by a distance of \(1 \mathrm{~m}\). What is the intensity of gravitational field at the mid point of the line joining them? (a) \(6.6 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\) (b) \(2.4 \times 10^{-8} \mathrm{Nkg}^{-1}\) (c) \(2.4 \times 10^{-7} \mathrm{Nkg}^{-1}\) (d) \(2.4 \times 10^{-6} \mathrm{Nkg}^{-1}\)

The height at which the acceleration due to gravity becomes \(g / 9\) (where \(g\) is acceleration due to gravity on the surface of the earth) in terms of \(R\), (the radius of the earth) is (a) \(R / \sqrt{2}\) (b) \(R / 2\) (c) \(\sqrt{2} R\) (d) \(2 R\)

There are two bodies of masses \(100000 \mathrm{~kg}\) and \(1000 \mathrm{~kg}\) separated by a distance of \(1 \mathrm{~m} .\) At what distance (in metre) from the smaller body, the intensity of gravitational field will be zero? (a) \(1 / 9\) (b) \(1 / 10\) (c) \(1 / 11\) (d) \(10 / 11\)

The effect of rotation of the earth on the value of acceleration due to gravity is (a) \(g\) is maximum at the equator and minimum at the poles (b) \(g\) is minimum at the equator and maximum at the poles (c) \(g\) is maximum at both places (d) \(g\) is minimum at both places

In a certain region of space, the gravitational field is given by \(-k / r\), where \(r\) is the distance and \(k\) is a constant. If the gravitational potential at \(r=r_{0}\) be \(V_{0}\), then what is the expression for the gravitational potential \(V\) ? (a) \(k \log \left(\frac{r}{r_{0}}\right)\) (b) \(k \log \left(\frac{r_{0}}{r}\right)\) (c) \(V_{0}+k \log \left(\frac{r}{r_{0}}\right)\) (d) \(V_{0}+k \log \left(\frac{r_{0}}{r}\right)\)

The escape velocity from the earth is \(11 \mathrm{kms}^{-1}\).The escape velocity from a planet having twice the radius and same mean density as that of earth is [UP SEE 2009] (a) \(5.5 \mathrm{kms}^{-1}\) (b) \(11 \mathrm{kms}^{-1}\) (c) \(22 \mathrm{kms}^{-1}\) (d) None of these

Angular momentum in absence of any external torque remains constant. If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e., \(\tau=0\) $$ \begin{aligned} &\therefore \quad \frac{d L}{d t}=0 \\ &\Rightarrow \quad l_{1} \omega_{1}=I_{2} \omega_{2} \\ &\text { Here, } M=2 \mathrm{~kg}, m=0.25 \mathrm{~kg}, r=0.2 \mathrm{~m} \\ &\omega_{1}=30 \mathrm{rad} \mathrm{s}^{-1} \end{aligned} $$ Hence, we get after putting the given values in Eq. (i) $$\frac{1}{2} \times 2 \times(0.2)^{2} \times 30=\frac{1}{2} \times(2+2 \times 0.25)(0.2)^{2} \times \omega_{2}$$ $$\begin{array}{ll}\Rightarrow & 1.2=0.05 \omega_{2} \\ \therefore & \omega_{2}=24 \mathrm{rad} \mathrm{s}^{-1}\end{array}$$

If \(g\) is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass \(m\) raised from the surface of the earth to a height equal to the radius \(R\) of the earth is [UP SEE 2008] (a) \(2 \mathrm{mgR}\) (b) \(m g R\) (c) \(\frac{1}{2} m g R\) (d) \(\frac{1}{4} m g R\)

Since, rod is bent at the middle, so each part of it will ha same length \(\left(\frac{L}{2}\right)\) and \(\operatorname{mass}\left(\frac{M}{2}\right)\) as shown Moment of inertia of each part about an axis passing through its one end $$ =\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2} $$ Hence, net moment of inertia about an axis passing through its middle point \(O\) is $$ I=\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2}+\frac{1}{3}\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^{2}=\frac{1}{3}\left[\frac{M L^{2}}{8}+\frac{M L^{2}}{8}\right]=\frac{M L^{2}}{12} $$

A solid sphere is of density \(\rho\) and radius \(R\). The gravitational field at a distance \(r\) from the centre of the sphere, where \(r

If the distance between the sun and the earth is increased by three times, then attraction between two will (a) remain constant (b) decrease by \(63 \%\) (c) increase by \(63 \%\) (d) decrease by \(89 \%\)

Two bodies of masses \(2 \mathrm{~kg}\) and \(8 \mathrm{~kg}\) are separated by a distance of \(9 \mathrm{~m}\). The point where the resultant gravitational field intensity is zero is at a distance of (a) \(4.5 \mathrm{~m}\) from each mass (b) \(6 \mathrm{~m}\) from \(2 \mathrm{~kg}\) (c) \(6 \mathrm{~m}\) from \(8 \mathrm{~kg}\) (d) \(2.5 \mathrm{~m}\) from \(2 \mathrm{~kg}\)

The orbit of geostationary satellite is circular, the time period of satellite depends on (i) mass of the satellite, (ii) mass of the earth, (iii) radius of the orbit, (iv) height of the satellite form the surface of the earth. (a) (i) only (b) (i) and (ii) (c) (i), (ii) and (iii) (d) (ii), (iii) and (iv)

Here, \(\mathbf{F}=-F \hat{\mathbf{k}}\) $$ \begin{aligned} \mathbf{r} &=(\hat{\mathbf{i}}-\hat{\mathbf{j}}), \tau=? \\\ \text { As } & \tau=\mathbf{r} \times \mathbf{F}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \times(-F \hat{\mathbf{k}}) \end{aligned} $$ \(=-F(\hat{\mathbf{i}} \times \hat{\mathbf{k}}-\hat{\mathbf{j}} \times \hat{\mathbf{k}})\) \(=-F(-\hat{\mathbf{j}}-\hat{\mathbf{i}})\) \(=F(\hat{\mathbf{i}}+\hat{\mathbf{j}})\)

The weight of a body on surface of earth is \(121.6 \mathrm{~N}\). When it is raised to a height half the radius of earth, its weight will be (a) \(2.8 \mathrm{~N}\) (b) \(5.6 \mathrm{~N}\) (c) \(12.6 \mathrm{~N}\) (d) \(25.2 \mathrm{~N}\)

A particle of mass \(m\) is placed at the centre of a uniform spherical shell of mass \(3 \mathrm{~m}\) and radius \(R\). The gravitational potential on the surface of the shell is (a) \(-\frac{G m}{R}\) (b) \(-\frac{3 \mathrm{Gm}}{R}\) (c) \(-\frac{4 G m}{R}\) (d) \(-\frac{2 G m}{R}\)