Question 121:
easyThe weight of a body at a distance \(r\) (\(r > R\)) from the centre of earth is \(W\). The weight at a distance \(2r\) from the centre of earth is : (\(R \rightarrow\) Radius of earth)
Outside the earth, weight \(W \propto \frac{1}{r^2}\). When distance is doubled to \(2r\), the weight becomes \(\frac{1}{2^2} = \frac{1}{4}\) of its initial value, i.e., \(\frac{W}{4}\).
Question 122:
easyIf mass of a planet is \(M\) and radius is \(x\), then the work to be done to slowly take a mass \(m\) from surface of planet to a height \(4x\) is :
Initial position \(r_i = x\), final position \(r_f = x + 4x = 5x\). Work done \(W = U_f - U_i = -\frac{GMm}{5x} - \left(-\frac{GMm}{x}\right) = \frac{4GMm}{5x}\).
Question 123:
easyA planet of mass \(m\) is moving in an elliptical orbit about the sun with time period \(T\). If \(A\) be the area of orbit, then its angular momentum would be :
According to Kepler's second law, the areal velocity of the planet is constant and is given by \(\frac{dA}{dt} = \frac{L}{2m}\). Integrating over one time period \(T\) gives \(A = \frac{L}{2m} T ⇒ L = \frac{2mA}{T}\).
Question 124:
easyFor energy of satellite, match the columns (symbols have their respective meaning):
**Column-I**
(i) Kinetic energy
(ii) Potential energy
(iii) Total energy
**Column-II**
(p) \(\frac{L^2}{2mr^2}\)
(q) \(-\frac{L^2}{mr^2}\)
(r) \(-\frac{L^2}{2mr^2}\)
Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{L^2}{2mr^2}\) (since \(L = mvr\)). Potential energy \(U = -\frac{GMm}{r} = -\frac{L^2}{mr^2}\). Total energy \(E = K + U = -\frac{L^2}{2mr^2}\). Thus (i)-p, (ii)-q, (iii)-r.
Question 125:
easyAssertion : If rotation of earth about its own axis is suddenly stopped then acceleration due to gravity will increase at all places on the earth (except poles).
Reason : At height \(h\) from the surface of earth, acceleration due to gravity is \(g_h = g \left(1 – \frac{2h}{R_e}\right)\) (If \(h \ll R_e\)) [\(R_e \rightarrow\) radius of earth]
The effective gravity is \(g' = g - \omega^2 R \cos^2\lambda\). If rotation stops (\(\omega = 0\)), \(g'\) increases everywhere except the poles (where \(\lambda = 90^\circ\)). Thus Assertion is true. The Reason is also a true independent formula for gravity at a height, but not the explanation.
Question 126:
easyThe gravitational field in a region is given by \(vec{I} = 10(\hat{i} + \hat{j})\text{ N/kg}\). The work done by gravitational field to shift a particle of mass \(2\text{ kg}\) from position \((0,0)\) to \((5, 4)\) will be :
The force is \(\vec{F} = mvec{I} = 2 \times 10(\hat{i} + \hat{j}) = 20\hat{i} + 20\hat{j}\text{ N}\). Displacement is \(\vec{d} = 5\hat{i} + 4\hat{j}\text{ m}\). Work done \(W = \vec{F} \cdot \vec{d} = (20)(5) + (20)(4) = 180\text{ J}\).
Question 127:
moderateThe energy required to put a satellite of mass \(m\) from earth surface into a orbit of radius \(2R\) is \(E_1\). The energy further needed to change the orbit of this satellite from its present orbit to radius \(4R\) is \(E_2\). The ratio \(\frac{E_1}{E_2}\) is (where \(R\) is radius of earth:
The energy required to put a satellite in orbit from earth's surface is \(E_1 = -\frac{GMm}{2(2R)} - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R}\). The energy to change orbit from \(2R\) to \(4R\) is \(E_2 = -\frac{GMm}{2(4R)} - \left(-\frac{GMm}{2(2R)}\right) = \frac{GMm}{8R}\). Thus, \(\frac{E_1}{E_2} = 6\), which gives the ratio \(6 : 1\).
Question 128:
easyA body weighs \(900\text{ N}\) on the surface of earth. How much will it weigh at a height double the radius of earth?
Using the formula for acceleration due to gravity at height \(h\): \(g' = g\left(\frac{R}{R+h}\right)^2\). For \(h = 2R\), \(g' = \frac{g}{9}\) . Hence, weight at this height is \(W' = \frac{W}{9} = \frac{900\text{ N}}{9} = 100\text{ N}\).
Question 129:
easyThe escape velocity from the Earth’s surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is
Escape velocity is given by \(v_e = R\sqrt{\frac{8\pi G\rho}{3}}\). Since the mass density \(rho\) is the same, \(v_e\) is directly proportional to radius \(R\). Therefore, \(v_e' = 4v\).
Question 130:
easyA particle of mass \(m\) is projected with a velocity \(v = k V_e\) (\(k < 1\)) from the surface of the earth. (\(V_e = \text{escape velocity}\)) The maximum height above the surface reached by the particle is
By conservation of mechanical energy: \(-\frac{GMm}{R} + \frac{1}{2}mv^2 = -\frac{GMm}{R+h}\). Since \(v = k \sqrt{\frac{2GM}{R}}\), we substitute to get \(-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}\), which yields \(h = \frac{R k^2}{1-k^2}\).