359645
Which of the following statement is correct?
1 Acceleration due to gravity increase with increasing altitude.
2 Acceleration due to gravity increase with increasing depth.
3 Acceleration due to gravity increase with increasing latitude.
4 Acceleration due to gravity is independent of the mass of the earth.
Explanation:
Acceleration due to gravity at an altitude \(h\) above the earth's surface is \(g_{b}=\dfrac{g R_{E}^{2}}{\left(R_{E}+h\right)^{2}}\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R_{E}\) is the radius of the earth. So acceleration due to gravity decreases with increasing altitude. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)\) So acceleration due to gravity decreases with increasing depth. Acceleration due to gravity at latitude \(\lambda\) \(g_{\lambda}=g-R_{E} \omega^{2} \cos ^{2} \lambda\) So acceleration due to gravity increases with increasing latitude \((\lambda)\). Acceleration due to gravity of body of mass \(m\) is placed on the earth's surface is \(g=\dfrac{G M_{E}}{R_{E}^{2}}\) So acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
PHXI08:GRAVITATION
359646
A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel after about
1 42 minute
2 84 minute
3 One day
4 One hour
Explanation:
Let \({A B}\) be the imaginary tunnel dug through the centre of the earth. A ball is released from \({A}\) in it. Consider that the density of earth be \({\rho}\) and radius of earth be \({R}\). As the ball released, for half journey \({A O}\) the acceleration due to gravity decreases. At a distance \({x}\) from centre, \({F=G \dfrac{4}{3} \pi x \rho m}\) acting towards \({O}\). \({\Rightarrow}\) Acceleration \({\propto x}\) The ball will execute \(SHM\) about the centre of the earth. The time taken to reach the other end \({=\dfrac{1}{2} \times}\) Time period of \(SHM\) \(\begin{aligned}t=\dfrac{1}{2} \times 2 \pi \sqrt{\dfrac{\text { displacement }}{\text { acceleration }}} & =\pi \sqrt{\dfrac{x}{(4 \pi / 3) G \rho x}} \\& =\sqrt{\dfrac{3 \pi}{4 G \rho}}\end{aligned}\) As \({G=6.67 \times 10^{-11} {~N} {~m}^{2} / {kg}^{2}}\) and \({\rho=5.52 \times 10^{3} {~kg} / {m}^{3}}\) \({t=\sqrt{\dfrac{3 \times 3.14}{4 \times 6.67 \times 10^{-11} \times 5.52 \times 10^{3}}}}\) \({=2529}\) seconds \({=42}\) minute 9 seconds. So correct option is (1)
PHXI08:GRAVITATION
359647
Find gravitational field at a distance of \(2000\;km\) from the centre of earth. (Given \({R_{earth{\rm{ }}}} = 6400\;km,r = 2000\;km,\) \(\left. {{M_{earth{\rm{ }}}} = 6 \times {{10}^{24}}\;kg} \right)\)
359648
The value of acceleration due to gravity at a height of 10 \(km\) from the surface of earth is \(x\). At what depth inside the earth is the value of the acceleration due to gravity has the same value \(x\) ?
359645
Which of the following statement is correct?
1 Acceleration due to gravity increase with increasing altitude.
2 Acceleration due to gravity increase with increasing depth.
3 Acceleration due to gravity increase with increasing latitude.
4 Acceleration due to gravity is independent of the mass of the earth.
Explanation:
Acceleration due to gravity at an altitude \(h\) above the earth's surface is \(g_{b}=\dfrac{g R_{E}^{2}}{\left(R_{E}+h\right)^{2}}\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R_{E}\) is the radius of the earth. So acceleration due to gravity decreases with increasing altitude. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)\) So acceleration due to gravity decreases with increasing depth. Acceleration due to gravity at latitude \(\lambda\) \(g_{\lambda}=g-R_{E} \omega^{2} \cos ^{2} \lambda\) So acceleration due to gravity increases with increasing latitude \((\lambda)\). Acceleration due to gravity of body of mass \(m\) is placed on the earth's surface is \(g=\dfrac{G M_{E}}{R_{E}^{2}}\) So acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
PHXI08:GRAVITATION
359646
A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel after about
1 42 minute
2 84 minute
3 One day
4 One hour
Explanation:
Let \({A B}\) be the imaginary tunnel dug through the centre of the earth. A ball is released from \({A}\) in it. Consider that the density of earth be \({\rho}\) and radius of earth be \({R}\). As the ball released, for half journey \({A O}\) the acceleration due to gravity decreases. At a distance \({x}\) from centre, \({F=G \dfrac{4}{3} \pi x \rho m}\) acting towards \({O}\). \({\Rightarrow}\) Acceleration \({\propto x}\) The ball will execute \(SHM\) about the centre of the earth. The time taken to reach the other end \({=\dfrac{1}{2} \times}\) Time period of \(SHM\) \(\begin{aligned}t=\dfrac{1}{2} \times 2 \pi \sqrt{\dfrac{\text { displacement }}{\text { acceleration }}} & =\pi \sqrt{\dfrac{x}{(4 \pi / 3) G \rho x}} \\& =\sqrt{\dfrac{3 \pi}{4 G \rho}}\end{aligned}\) As \({G=6.67 \times 10^{-11} {~N} {~m}^{2} / {kg}^{2}}\) and \({\rho=5.52 \times 10^{3} {~kg} / {m}^{3}}\) \({t=\sqrt{\dfrac{3 \times 3.14}{4 \times 6.67 \times 10^{-11} \times 5.52 \times 10^{3}}}}\) \({=2529}\) seconds \({=42}\) minute 9 seconds. So correct option is (1)
PHXI08:GRAVITATION
359647
Find gravitational field at a distance of \(2000\;km\) from the centre of earth. (Given \({R_{earth{\rm{ }}}} = 6400\;km,r = 2000\;km,\) \(\left. {{M_{earth{\rm{ }}}} = 6 \times {{10}^{24}}\;kg} \right)\)
359648
The value of acceleration due to gravity at a height of 10 \(km\) from the surface of earth is \(x\). At what depth inside the earth is the value of the acceleration due to gravity has the same value \(x\) ?
359645
Which of the following statement is correct?
1 Acceleration due to gravity increase with increasing altitude.
2 Acceleration due to gravity increase with increasing depth.
3 Acceleration due to gravity increase with increasing latitude.
4 Acceleration due to gravity is independent of the mass of the earth.
Explanation:
Acceleration due to gravity at an altitude \(h\) above the earth's surface is \(g_{b}=\dfrac{g R_{E}^{2}}{\left(R_{E}+h\right)^{2}}\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R_{E}\) is the radius of the earth. So acceleration due to gravity decreases with increasing altitude. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)\) So acceleration due to gravity decreases with increasing depth. Acceleration due to gravity at latitude \(\lambda\) \(g_{\lambda}=g-R_{E} \omega^{2} \cos ^{2} \lambda\) So acceleration due to gravity increases with increasing latitude \((\lambda)\). Acceleration due to gravity of body of mass \(m\) is placed on the earth's surface is \(g=\dfrac{G M_{E}}{R_{E}^{2}}\) So acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
PHXI08:GRAVITATION
359646
A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel after about
1 42 minute
2 84 minute
3 One day
4 One hour
Explanation:
Let \({A B}\) be the imaginary tunnel dug through the centre of the earth. A ball is released from \({A}\) in it. Consider that the density of earth be \({\rho}\) and radius of earth be \({R}\). As the ball released, for half journey \({A O}\) the acceleration due to gravity decreases. At a distance \({x}\) from centre, \({F=G \dfrac{4}{3} \pi x \rho m}\) acting towards \({O}\). \({\Rightarrow}\) Acceleration \({\propto x}\) The ball will execute \(SHM\) about the centre of the earth. The time taken to reach the other end \({=\dfrac{1}{2} \times}\) Time period of \(SHM\) \(\begin{aligned}t=\dfrac{1}{2} \times 2 \pi \sqrt{\dfrac{\text { displacement }}{\text { acceleration }}} & =\pi \sqrt{\dfrac{x}{(4 \pi / 3) G \rho x}} \\& =\sqrt{\dfrac{3 \pi}{4 G \rho}}\end{aligned}\) As \({G=6.67 \times 10^{-11} {~N} {~m}^{2} / {kg}^{2}}\) and \({\rho=5.52 \times 10^{3} {~kg} / {m}^{3}}\) \({t=\sqrt{\dfrac{3 \times 3.14}{4 \times 6.67 \times 10^{-11} \times 5.52 \times 10^{3}}}}\) \({=2529}\) seconds \({=42}\) minute 9 seconds. So correct option is (1)
PHXI08:GRAVITATION
359647
Find gravitational field at a distance of \(2000\;km\) from the centre of earth. (Given \({R_{earth{\rm{ }}}} = 6400\;km,r = 2000\;km,\) \(\left. {{M_{earth{\rm{ }}}} = 6 \times {{10}^{24}}\;kg} \right)\)
359648
The value of acceleration due to gravity at a height of 10 \(km\) from the surface of earth is \(x\). At what depth inside the earth is the value of the acceleration due to gravity has the same value \(x\) ?
359645
Which of the following statement is correct?
1 Acceleration due to gravity increase with increasing altitude.
2 Acceleration due to gravity increase with increasing depth.
3 Acceleration due to gravity increase with increasing latitude.
4 Acceleration due to gravity is independent of the mass of the earth.
Explanation:
Acceleration due to gravity at an altitude \(h\) above the earth's surface is \(g_{b}=\dfrac{g R_{E}^{2}}{\left(R_{E}+h\right)^{2}}\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R_{E}\) is the radius of the earth. So acceleration due to gravity decreases with increasing altitude. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)\) So acceleration due to gravity decreases with increasing depth. Acceleration due to gravity at latitude \(\lambda\) \(g_{\lambda}=g-R_{E} \omega^{2} \cos ^{2} \lambda\) So acceleration due to gravity increases with increasing latitude \((\lambda)\). Acceleration due to gravity of body of mass \(m\) is placed on the earth's surface is \(g=\dfrac{G M_{E}}{R_{E}^{2}}\) So acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
PHXI08:GRAVITATION
359646
A tunnel has been dug through the centre of the earth and a ball is released in it. It will reach the other end of the tunnel after about
1 42 minute
2 84 minute
3 One day
4 One hour
Explanation:
Let \({A B}\) be the imaginary tunnel dug through the centre of the earth. A ball is released from \({A}\) in it. Consider that the density of earth be \({\rho}\) and radius of earth be \({R}\). As the ball released, for half journey \({A O}\) the acceleration due to gravity decreases. At a distance \({x}\) from centre, \({F=G \dfrac{4}{3} \pi x \rho m}\) acting towards \({O}\). \({\Rightarrow}\) Acceleration \({\propto x}\) The ball will execute \(SHM\) about the centre of the earth. The time taken to reach the other end \({=\dfrac{1}{2} \times}\) Time period of \(SHM\) \(\begin{aligned}t=\dfrac{1}{2} \times 2 \pi \sqrt{\dfrac{\text { displacement }}{\text { acceleration }}} & =\pi \sqrt{\dfrac{x}{(4 \pi / 3) G \rho x}} \\& =\sqrt{\dfrac{3 \pi}{4 G \rho}}\end{aligned}\) As \({G=6.67 \times 10^{-11} {~N} {~m}^{2} / {kg}^{2}}\) and \({\rho=5.52 \times 10^{3} {~kg} / {m}^{3}}\) \({t=\sqrt{\dfrac{3 \times 3.14}{4 \times 6.67 \times 10^{-11} \times 5.52 \times 10^{3}}}}\) \({=2529}\) seconds \({=42}\) minute 9 seconds. So correct option is (1)
PHXI08:GRAVITATION
359647
Find gravitational field at a distance of \(2000\;km\) from the centre of earth. (Given \({R_{earth{\rm{ }}}} = 6400\;km,r = 2000\;km,\) \(\left. {{M_{earth{\rm{ }}}} = 6 \times {{10}^{24}}\;kg} \right)\)
359648
The value of acceleration due to gravity at a height of 10 \(km\) from the surface of earth is \(x\). At what depth inside the earth is the value of the acceleration due to gravity has the same value \(x\) ?
