359662
Let \(g_{h}\) and \(g_{d}\) be the acceleration due to gravity at height \(h\) above the earth's surface and at depth \(d\) below the earth's surface respectively. If \(g_{h}=g_{d}\), then the relation between \(h\) and \(d\) is
1 \(d=h\)
2 \(d=\dfrac{h}{2}\)
3 \(d=\dfrac{h}{4}\)
4 \(d=2 h\)
Explanation:
Acceleration due to gravity at a height \(h\) above the earth's surface \(g_{h}=g\left(1-\dfrac{2 h}{R}\right)\) Acceleration due to gravity at a depth \(d\) below the earth's surface \(g_{d}=g\left(1-\dfrac{d}{R}\right)\) \(\begin{aligned}& \text { As } g_{h}=g_{d} \\& \therefore g\left(1-\dfrac{2 h}{R}\right)=g\left(1-\dfrac{d}{R}\right) \\& \Rightarrow d=2 h\end{aligned}\)
MHTCET - 2015
PHXI08:GRAVITATION
359663
At what distance above and below the surface of the earth, a body will have same weight. (Take radius of earth as \(R\) )
1 \(\dfrac{R}{2}\)
2 \(\sqrt{5} R-R\)
3 \(\dfrac{\sqrt{3} R-R}{2}\)
4 \(\dfrac{\sqrt{5} R-R}{2}\)
Explanation:
Weight of body \(=m g\) \(\quad \quad (1)\) Acceleration due to gravity at height \(h\) and depth \(h\) is given by \(g\) at height \(h=\dfrac{g R^{2}}{(R+h)^{2}}\) \(\quad \quad (2)\) \(g\) at depth \(h=g\left(1-\dfrac{h}{R}\right)\) \(\quad \quad (3)\) Given, same weight at height \(h\) and depth \(h\) Using equation (1), (2) and (3) \(\begin{aligned}& \Rightarrow \dfrac{m g R^{2}}{(R+h)^{2}}=m g\left(1-\dfrac{h}{R}\right) ; \dfrac{R^{2}}{(R+h)^{2}}=1-\dfrac{h}{R} \\& \Rightarrow R^{2}=(R+h)^{2}\left(1-\dfrac{h}{R}\right) ; 1 \\& =\left(\dfrac{R^{2}+h^{2}+2 R h}{R^{2}}\right)\left(1-\dfrac{h}{R}\right) \\& \Rightarrow\left(1+\dfrac{h^{2}}{R^{2}}+\dfrac{2 h}{R}\right)\left(1-\dfrac{h}{R}\right)=1 \\& \Rightarrow 1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-\dfrac{h^{3}}{R^{3}}+\dfrac{2 h}{R}-\dfrac{2 h^{2}}{R^{2}}=1 \\& \Rightarrow-\dfrac{h}{R}\left(1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-2+\dfrac{2 h}{R}\right)=0 \\& \Rightarrow-1+\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}=0 ; \\& \Rightarrow h^{2}+h R-R^{2}=0\end{aligned}\) After solving, we get, \(h=\dfrac{\sqrt{5} R-R}{2}\)
JEE - 2024
PHXI08:GRAVITATION
359664
The depth at which the value of acceleration due to gravity becomes \(\dfrac{1}{n}\) times the value at the surface is
1 \(\dfrac{R}{n}\)
2 \(\dfrac{R}{n^{2}}\)
3 \(\dfrac{R[n-1]}{n}\)
4 \(\dfrac{R n}{n-1}\)
Explanation:
At depth \(d\) from earth surface \(g^{\prime}=g\left[1-\dfrac{d}{R}\right]\) \(g^{\prime}=\dfrac{g}{n}\) \(\dfrac{g}{n}=g\left[1-\dfrac{d}{R}\right]\) \(\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\) \(d=\left(\dfrac{n-1}{n}\right) R\).
PHXI08:GRAVITATION
359665
The ratio between the values of acceleration due to gravity at a height 2 \(km\) above and at a depth of 1\(km\) below the earth's surface is (radius of earth is \(R\) ):
1 \(\dfrac{R-4}{R-2}\)
2 \(\dfrac{R}{R-1}\)
3 \(\dfrac{R-2}{R}\)
4 \(\dfrac{2 R}{R-1}\)
Explanation:
Here, \(h = 2\;km,\;d = 2\;km\) Acceleration due to gravity at a height \(h\) above the earth's surface is \({g_h} = g\left( {1 - \frac{{2\;h}}{R}} \right)\) for \(h < < R\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R\) is the radius of the earth. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(\begin{aligned}& \mathrm{g}_{d}=g\left(1-\dfrac{d}{R}\right) \\& \therefore \dfrac{g_{h}}{g_{d}}=\dfrac{\left(1-\dfrac{2 h}{R}\right)}{\left(1-\dfrac{d}{R}\right)}=\dfrac{R-2 h}{(R-d)}=\dfrac{R-4}{R-2}\end{aligned}\)
359662
Let \(g_{h}\) and \(g_{d}\) be the acceleration due to gravity at height \(h\) above the earth's surface and at depth \(d\) below the earth's surface respectively. If \(g_{h}=g_{d}\), then the relation between \(h\) and \(d\) is
1 \(d=h\)
2 \(d=\dfrac{h}{2}\)
3 \(d=\dfrac{h}{4}\)
4 \(d=2 h\)
Explanation:
Acceleration due to gravity at a height \(h\) above the earth's surface \(g_{h}=g\left(1-\dfrac{2 h}{R}\right)\) Acceleration due to gravity at a depth \(d\) below the earth's surface \(g_{d}=g\left(1-\dfrac{d}{R}\right)\) \(\begin{aligned}& \text { As } g_{h}=g_{d} \\& \therefore g\left(1-\dfrac{2 h}{R}\right)=g\left(1-\dfrac{d}{R}\right) \\& \Rightarrow d=2 h\end{aligned}\)
MHTCET - 2015
PHXI08:GRAVITATION
359663
At what distance above and below the surface of the earth, a body will have same weight. (Take radius of earth as \(R\) )
1 \(\dfrac{R}{2}\)
2 \(\sqrt{5} R-R\)
3 \(\dfrac{\sqrt{3} R-R}{2}\)
4 \(\dfrac{\sqrt{5} R-R}{2}\)
Explanation:
Weight of body \(=m g\) \(\quad \quad (1)\) Acceleration due to gravity at height \(h\) and depth \(h\) is given by \(g\) at height \(h=\dfrac{g R^{2}}{(R+h)^{2}}\) \(\quad \quad (2)\) \(g\) at depth \(h=g\left(1-\dfrac{h}{R}\right)\) \(\quad \quad (3)\) Given, same weight at height \(h\) and depth \(h\) Using equation (1), (2) and (3) \(\begin{aligned}& \Rightarrow \dfrac{m g R^{2}}{(R+h)^{2}}=m g\left(1-\dfrac{h}{R}\right) ; \dfrac{R^{2}}{(R+h)^{2}}=1-\dfrac{h}{R} \\& \Rightarrow R^{2}=(R+h)^{2}\left(1-\dfrac{h}{R}\right) ; 1 \\& =\left(\dfrac{R^{2}+h^{2}+2 R h}{R^{2}}\right)\left(1-\dfrac{h}{R}\right) \\& \Rightarrow\left(1+\dfrac{h^{2}}{R^{2}}+\dfrac{2 h}{R}\right)\left(1-\dfrac{h}{R}\right)=1 \\& \Rightarrow 1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-\dfrac{h^{3}}{R^{3}}+\dfrac{2 h}{R}-\dfrac{2 h^{2}}{R^{2}}=1 \\& \Rightarrow-\dfrac{h}{R}\left(1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-2+\dfrac{2 h}{R}\right)=0 \\& \Rightarrow-1+\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}=0 ; \\& \Rightarrow h^{2}+h R-R^{2}=0\end{aligned}\) After solving, we get, \(h=\dfrac{\sqrt{5} R-R}{2}\)
JEE - 2024
PHXI08:GRAVITATION
359664
The depth at which the value of acceleration due to gravity becomes \(\dfrac{1}{n}\) times the value at the surface is
1 \(\dfrac{R}{n}\)
2 \(\dfrac{R}{n^{2}}\)
3 \(\dfrac{R[n-1]}{n}\)
4 \(\dfrac{R n}{n-1}\)
Explanation:
At depth \(d\) from earth surface \(g^{\prime}=g\left[1-\dfrac{d}{R}\right]\) \(g^{\prime}=\dfrac{g}{n}\) \(\dfrac{g}{n}=g\left[1-\dfrac{d}{R}\right]\) \(\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\) \(d=\left(\dfrac{n-1}{n}\right) R\).
PHXI08:GRAVITATION
359665
The ratio between the values of acceleration due to gravity at a height 2 \(km\) above and at a depth of 1\(km\) below the earth's surface is (radius of earth is \(R\) ):
1 \(\dfrac{R-4}{R-2}\)
2 \(\dfrac{R}{R-1}\)
3 \(\dfrac{R-2}{R}\)
4 \(\dfrac{2 R}{R-1}\)
Explanation:
Here, \(h = 2\;km,\;d = 2\;km\) Acceleration due to gravity at a height \(h\) above the earth's surface is \({g_h} = g\left( {1 - \frac{{2\;h}}{R}} \right)\) for \(h < < R\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R\) is the radius of the earth. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(\begin{aligned}& \mathrm{g}_{d}=g\left(1-\dfrac{d}{R}\right) \\& \therefore \dfrac{g_{h}}{g_{d}}=\dfrac{\left(1-\dfrac{2 h}{R}\right)}{\left(1-\dfrac{d}{R}\right)}=\dfrac{R-2 h}{(R-d)}=\dfrac{R-4}{R-2}\end{aligned}\)
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PHXI08:GRAVITATION
359662
Let \(g_{h}\) and \(g_{d}\) be the acceleration due to gravity at height \(h\) above the earth's surface and at depth \(d\) below the earth's surface respectively. If \(g_{h}=g_{d}\), then the relation between \(h\) and \(d\) is
1 \(d=h\)
2 \(d=\dfrac{h}{2}\)
3 \(d=\dfrac{h}{4}\)
4 \(d=2 h\)
Explanation:
Acceleration due to gravity at a height \(h\) above the earth's surface \(g_{h}=g\left(1-\dfrac{2 h}{R}\right)\) Acceleration due to gravity at a depth \(d\) below the earth's surface \(g_{d}=g\left(1-\dfrac{d}{R}\right)\) \(\begin{aligned}& \text { As } g_{h}=g_{d} \\& \therefore g\left(1-\dfrac{2 h}{R}\right)=g\left(1-\dfrac{d}{R}\right) \\& \Rightarrow d=2 h\end{aligned}\)
MHTCET - 2015
PHXI08:GRAVITATION
359663
At what distance above and below the surface of the earth, a body will have same weight. (Take radius of earth as \(R\) )
1 \(\dfrac{R}{2}\)
2 \(\sqrt{5} R-R\)
3 \(\dfrac{\sqrt{3} R-R}{2}\)
4 \(\dfrac{\sqrt{5} R-R}{2}\)
Explanation:
Weight of body \(=m g\) \(\quad \quad (1)\) Acceleration due to gravity at height \(h\) and depth \(h\) is given by \(g\) at height \(h=\dfrac{g R^{2}}{(R+h)^{2}}\) \(\quad \quad (2)\) \(g\) at depth \(h=g\left(1-\dfrac{h}{R}\right)\) \(\quad \quad (3)\) Given, same weight at height \(h\) and depth \(h\) Using equation (1), (2) and (3) \(\begin{aligned}& \Rightarrow \dfrac{m g R^{2}}{(R+h)^{2}}=m g\left(1-\dfrac{h}{R}\right) ; \dfrac{R^{2}}{(R+h)^{2}}=1-\dfrac{h}{R} \\& \Rightarrow R^{2}=(R+h)^{2}\left(1-\dfrac{h}{R}\right) ; 1 \\& =\left(\dfrac{R^{2}+h^{2}+2 R h}{R^{2}}\right)\left(1-\dfrac{h}{R}\right) \\& \Rightarrow\left(1+\dfrac{h^{2}}{R^{2}}+\dfrac{2 h}{R}\right)\left(1-\dfrac{h}{R}\right)=1 \\& \Rightarrow 1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-\dfrac{h^{3}}{R^{3}}+\dfrac{2 h}{R}-\dfrac{2 h^{2}}{R^{2}}=1 \\& \Rightarrow-\dfrac{h}{R}\left(1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-2+\dfrac{2 h}{R}\right)=0 \\& \Rightarrow-1+\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}=0 ; \\& \Rightarrow h^{2}+h R-R^{2}=0\end{aligned}\) After solving, we get, \(h=\dfrac{\sqrt{5} R-R}{2}\)
JEE - 2024
PHXI08:GRAVITATION
359664
The depth at which the value of acceleration due to gravity becomes \(\dfrac{1}{n}\) times the value at the surface is
1 \(\dfrac{R}{n}\)
2 \(\dfrac{R}{n^{2}}\)
3 \(\dfrac{R[n-1]}{n}\)
4 \(\dfrac{R n}{n-1}\)
Explanation:
At depth \(d\) from earth surface \(g^{\prime}=g\left[1-\dfrac{d}{R}\right]\) \(g^{\prime}=\dfrac{g}{n}\) \(\dfrac{g}{n}=g\left[1-\dfrac{d}{R}\right]\) \(\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\) \(d=\left(\dfrac{n-1}{n}\right) R\).
PHXI08:GRAVITATION
359665
The ratio between the values of acceleration due to gravity at a height 2 \(km\) above and at a depth of 1\(km\) below the earth's surface is (radius of earth is \(R\) ):
1 \(\dfrac{R-4}{R-2}\)
2 \(\dfrac{R}{R-1}\)
3 \(\dfrac{R-2}{R}\)
4 \(\dfrac{2 R}{R-1}\)
Explanation:
Here, \(h = 2\;km,\;d = 2\;km\) Acceleration due to gravity at a height \(h\) above the earth's surface is \({g_h} = g\left( {1 - \frac{{2\;h}}{R}} \right)\) for \(h < < R\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R\) is the radius of the earth. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(\begin{aligned}& \mathrm{g}_{d}=g\left(1-\dfrac{d}{R}\right) \\& \therefore \dfrac{g_{h}}{g_{d}}=\dfrac{\left(1-\dfrac{2 h}{R}\right)}{\left(1-\dfrac{d}{R}\right)}=\dfrac{R-2 h}{(R-d)}=\dfrac{R-4}{R-2}\end{aligned}\)
359662
Let \(g_{h}\) and \(g_{d}\) be the acceleration due to gravity at height \(h\) above the earth's surface and at depth \(d\) below the earth's surface respectively. If \(g_{h}=g_{d}\), then the relation between \(h\) and \(d\) is
1 \(d=h\)
2 \(d=\dfrac{h}{2}\)
3 \(d=\dfrac{h}{4}\)
4 \(d=2 h\)
Explanation:
Acceleration due to gravity at a height \(h\) above the earth's surface \(g_{h}=g\left(1-\dfrac{2 h}{R}\right)\) Acceleration due to gravity at a depth \(d\) below the earth's surface \(g_{d}=g\left(1-\dfrac{d}{R}\right)\) \(\begin{aligned}& \text { As } g_{h}=g_{d} \\& \therefore g\left(1-\dfrac{2 h}{R}\right)=g\left(1-\dfrac{d}{R}\right) \\& \Rightarrow d=2 h\end{aligned}\)
MHTCET - 2015
PHXI08:GRAVITATION
359663
At what distance above and below the surface of the earth, a body will have same weight. (Take radius of earth as \(R\) )
1 \(\dfrac{R}{2}\)
2 \(\sqrt{5} R-R\)
3 \(\dfrac{\sqrt{3} R-R}{2}\)
4 \(\dfrac{\sqrt{5} R-R}{2}\)
Explanation:
Weight of body \(=m g\) \(\quad \quad (1)\) Acceleration due to gravity at height \(h\) and depth \(h\) is given by \(g\) at height \(h=\dfrac{g R^{2}}{(R+h)^{2}}\) \(\quad \quad (2)\) \(g\) at depth \(h=g\left(1-\dfrac{h}{R}\right)\) \(\quad \quad (3)\) Given, same weight at height \(h\) and depth \(h\) Using equation (1), (2) and (3) \(\begin{aligned}& \Rightarrow \dfrac{m g R^{2}}{(R+h)^{2}}=m g\left(1-\dfrac{h}{R}\right) ; \dfrac{R^{2}}{(R+h)^{2}}=1-\dfrac{h}{R} \\& \Rightarrow R^{2}=(R+h)^{2}\left(1-\dfrac{h}{R}\right) ; 1 \\& =\left(\dfrac{R^{2}+h^{2}+2 R h}{R^{2}}\right)\left(1-\dfrac{h}{R}\right) \\& \Rightarrow\left(1+\dfrac{h^{2}}{R^{2}}+\dfrac{2 h}{R}\right)\left(1-\dfrac{h}{R}\right)=1 \\& \Rightarrow 1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-\dfrac{h^{3}}{R^{3}}+\dfrac{2 h}{R}-\dfrac{2 h^{2}}{R^{2}}=1 \\& \Rightarrow-\dfrac{h}{R}\left(1-\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}-2+\dfrac{2 h}{R}\right)=0 \\& \Rightarrow-1+\dfrac{h}{R}+\dfrac{h^{2}}{R^{2}}=0 ; \\& \Rightarrow h^{2}+h R-R^{2}=0\end{aligned}\) After solving, we get, \(h=\dfrac{\sqrt{5} R-R}{2}\)
JEE - 2024
PHXI08:GRAVITATION
359664
The depth at which the value of acceleration due to gravity becomes \(\dfrac{1}{n}\) times the value at the surface is
1 \(\dfrac{R}{n}\)
2 \(\dfrac{R}{n^{2}}\)
3 \(\dfrac{R[n-1]}{n}\)
4 \(\dfrac{R n}{n-1}\)
Explanation:
At depth \(d\) from earth surface \(g^{\prime}=g\left[1-\dfrac{d}{R}\right]\) \(g^{\prime}=\dfrac{g}{n}\) \(\dfrac{g}{n}=g\left[1-\dfrac{d}{R}\right]\) \(\dfrac{d}{R}=1-\dfrac{1}{n}=\dfrac{n-1}{n}\) \(d=\left(\dfrac{n-1}{n}\right) R\).
PHXI08:GRAVITATION
359665
The ratio between the values of acceleration due to gravity at a height 2 \(km\) above and at a depth of 1\(km\) below the earth's surface is (radius of earth is \(R\) ):
1 \(\dfrac{R-4}{R-2}\)
2 \(\dfrac{R}{R-1}\)
3 \(\dfrac{R-2}{R}\)
4 \(\dfrac{2 R}{R-1}\)
Explanation:
Here, \(h = 2\;km,\;d = 2\;km\) Acceleration due to gravity at a height \(h\) above the earth's surface is \({g_h} = g\left( {1 - \frac{{2\;h}}{R}} \right)\) for \(h < < R\) Where \(g\) is the acceleration due to gravity on the earth's surface and \(R\) is the radius of the earth. Acceleration due to gravity at a depth \(d\) below the earth's surface is \(\begin{aligned}& \mathrm{g}_{d}=g\left(1-\dfrac{d}{R}\right) \\& \therefore \dfrac{g_{h}}{g_{d}}=\dfrac{\left(1-\dfrac{2 h}{R}\right)}{\left(1-\dfrac{d}{R}\right)}=\dfrac{R-2 h}{(R-d)}=\dfrac{R-4}{R-2}\end{aligned}\)