138383
When you move from equator to poles, the value of acceleration due to gravity (g)
1 increases
2 decreases
3 remains the same
4 increases then decreases
Explanation:
A Value of $g$ varies with latitude $(\lambda)$ as $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2} \cos ^{2} \lambda$ $\therefore$ At poles $\lambda=90^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}$ (maximum) At equator $\lambda=0^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2}$ (minimum) Also, the distance between equations is greater than the distance between poles by $21 \mathrm{~km}$.
SRMJEE - 2013
Gravitation
138384
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:
A At ball will execute S.H.M. $\mathrm{T}=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{6400 \times 10^{3}}{9.8}} \times \frac{1}{60}$ $\mathrm{~T}=84.63$ So, time taken to reach the other end is $t=\frac{84.63}{2}$ $t=42.31 \mathrm{~min}$
SRMJEE - 2014
Gravitation
138385
The planets with radii $R_{1}$ and $R_{2}$ have densities $\rho_{1}, \rho_{2}$ respectively. Their atmospheric pressure are $p_{1}$ and $p_{2}$ respectively. Therefore the ratio of masses of their atmospheres, neglecting variation of $g$ within the limits of atmosphere is
D We know that, Acceleration due to gravity- $\mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{R}^{2}}$ $\mathrm{~g}=\frac{4}{3} \pi \rho \mathrm{GR}$ $\therefore \quad \frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\rho_{1} \mathrm{R}_{1}}{\rho_{2} \mathrm{R}_{2}}$ Pressure $=\frac{\mathrm{w}}{\mathrm{S}} \Rightarrow \mathrm{p}=\frac{\mathrm{mg}}{\mathrm{S}} \Rightarrow \mathrm{m}=\frac{\mathrm{pS}}{\mathrm{g}}$ Where $\mathrm{w}=$ weight of atmosphere. $\mathrm{S}=$ Surface area of the planet. $\frac{m_{1}}{m_{2}}=\frac{p_{1} S_{1} g_{2}}{p_{2} S_{2} g_{1}}$ $\frac{m_{1}}{m_{2}}=\frac{p_{1} \cdot\left(4 \pi R_{1}^{2}\right)}{p_{2}\left(4 \pi R_{2}^{2}\right)} \times \frac{\rho_{2} R_{2}}{\rho_{1} R_{1}}$\$\$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{p}_{1} \mathrm{R}_{1} \rho_{2}}{\mathrm{p}_{2} \mathrm{R}_{2} \rho_{1}}$
JIPMER-2017
Gravitation
138386
At what height above the earth's surface, the value of $g$ is same as in a mine $80 \mathrm{~km}$ deep?
1 $20 \mathrm{~km}$
2 $30 \mathrm{~km}$
3 $40 \mathrm{~km}$
4 $50 \mathrm{~km}$
Explanation:
C We know that, Variation of $\mathrm{g}$ with height $(\mathrm{h})$, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Variation of $\mathrm{g}$ with depth $(\mathrm{d})=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ Now, $g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$ $1-\frac{2 h}{R}=1-\frac{d}{R}$ $-\frac{2 h}{R}=-\frac{d}{R}$ $h=\frac{d}{2}$ $h=\frac{80}{2}$ $h=40 \mathrm{~km}$
138383
When you move from equator to poles, the value of acceleration due to gravity (g)
1 increases
2 decreases
3 remains the same
4 increases then decreases
Explanation:
A Value of $g$ varies with latitude $(\lambda)$ as $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2} \cos ^{2} \lambda$ $\therefore$ At poles $\lambda=90^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}$ (maximum) At equator $\lambda=0^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2}$ (minimum) Also, the distance between equations is greater than the distance between poles by $21 \mathrm{~km}$.
SRMJEE - 2013
Gravitation
138384
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:
A At ball will execute S.H.M. $\mathrm{T}=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{6400 \times 10^{3}}{9.8}} \times \frac{1}{60}$ $\mathrm{~T}=84.63$ So, time taken to reach the other end is $t=\frac{84.63}{2}$ $t=42.31 \mathrm{~min}$
SRMJEE - 2014
Gravitation
138385
The planets with radii $R_{1}$ and $R_{2}$ have densities $\rho_{1}, \rho_{2}$ respectively. Their atmospheric pressure are $p_{1}$ and $p_{2}$ respectively. Therefore the ratio of masses of their atmospheres, neglecting variation of $g$ within the limits of atmosphere is
D We know that, Acceleration due to gravity- $\mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{R}^{2}}$ $\mathrm{~g}=\frac{4}{3} \pi \rho \mathrm{GR}$ $\therefore \quad \frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\rho_{1} \mathrm{R}_{1}}{\rho_{2} \mathrm{R}_{2}}$ Pressure $=\frac{\mathrm{w}}{\mathrm{S}} \Rightarrow \mathrm{p}=\frac{\mathrm{mg}}{\mathrm{S}} \Rightarrow \mathrm{m}=\frac{\mathrm{pS}}{\mathrm{g}}$ Where $\mathrm{w}=$ weight of atmosphere. $\mathrm{S}=$ Surface area of the planet. $\frac{m_{1}}{m_{2}}=\frac{p_{1} S_{1} g_{2}}{p_{2} S_{2} g_{1}}$ $\frac{m_{1}}{m_{2}}=\frac{p_{1} \cdot\left(4 \pi R_{1}^{2}\right)}{p_{2}\left(4 \pi R_{2}^{2}\right)} \times \frac{\rho_{2} R_{2}}{\rho_{1} R_{1}}$\$\$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{p}_{1} \mathrm{R}_{1} \rho_{2}}{\mathrm{p}_{2} \mathrm{R}_{2} \rho_{1}}$
JIPMER-2017
Gravitation
138386
At what height above the earth's surface, the value of $g$ is same as in a mine $80 \mathrm{~km}$ deep?
1 $20 \mathrm{~km}$
2 $30 \mathrm{~km}$
3 $40 \mathrm{~km}$
4 $50 \mathrm{~km}$
Explanation:
C We know that, Variation of $\mathrm{g}$ with height $(\mathrm{h})$, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Variation of $\mathrm{g}$ with depth $(\mathrm{d})=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ Now, $g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$ $1-\frac{2 h}{R}=1-\frac{d}{R}$ $-\frac{2 h}{R}=-\frac{d}{R}$ $h=\frac{d}{2}$ $h=\frac{80}{2}$ $h=40 \mathrm{~km}$
138383
When you move from equator to poles, the value of acceleration due to gravity (g)
1 increases
2 decreases
3 remains the same
4 increases then decreases
Explanation:
A Value of $g$ varies with latitude $(\lambda)$ as $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2} \cos ^{2} \lambda$ $\therefore$ At poles $\lambda=90^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}$ (maximum) At equator $\lambda=0^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2}$ (minimum) Also, the distance between equations is greater than the distance between poles by $21 \mathrm{~km}$.
SRMJEE - 2013
Gravitation
138384
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:
A At ball will execute S.H.M. $\mathrm{T}=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{6400 \times 10^{3}}{9.8}} \times \frac{1}{60}$ $\mathrm{~T}=84.63$ So, time taken to reach the other end is $t=\frac{84.63}{2}$ $t=42.31 \mathrm{~min}$
SRMJEE - 2014
Gravitation
138385
The planets with radii $R_{1}$ and $R_{2}$ have densities $\rho_{1}, \rho_{2}$ respectively. Their atmospheric pressure are $p_{1}$ and $p_{2}$ respectively. Therefore the ratio of masses of their atmospheres, neglecting variation of $g$ within the limits of atmosphere is
D We know that, Acceleration due to gravity- $\mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{R}^{2}}$ $\mathrm{~g}=\frac{4}{3} \pi \rho \mathrm{GR}$ $\therefore \quad \frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\rho_{1} \mathrm{R}_{1}}{\rho_{2} \mathrm{R}_{2}}$ Pressure $=\frac{\mathrm{w}}{\mathrm{S}} \Rightarrow \mathrm{p}=\frac{\mathrm{mg}}{\mathrm{S}} \Rightarrow \mathrm{m}=\frac{\mathrm{pS}}{\mathrm{g}}$ Where $\mathrm{w}=$ weight of atmosphere. $\mathrm{S}=$ Surface area of the planet. $\frac{m_{1}}{m_{2}}=\frac{p_{1} S_{1} g_{2}}{p_{2} S_{2} g_{1}}$ $\frac{m_{1}}{m_{2}}=\frac{p_{1} \cdot\left(4 \pi R_{1}^{2}\right)}{p_{2}\left(4 \pi R_{2}^{2}\right)} \times \frac{\rho_{2} R_{2}}{\rho_{1} R_{1}}$\$\$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{p}_{1} \mathrm{R}_{1} \rho_{2}}{\mathrm{p}_{2} \mathrm{R}_{2} \rho_{1}}$
JIPMER-2017
Gravitation
138386
At what height above the earth's surface, the value of $g$ is same as in a mine $80 \mathrm{~km}$ deep?
1 $20 \mathrm{~km}$
2 $30 \mathrm{~km}$
3 $40 \mathrm{~km}$
4 $50 \mathrm{~km}$
Explanation:
C We know that, Variation of $\mathrm{g}$ with height $(\mathrm{h})$, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Variation of $\mathrm{g}$ with depth $(\mathrm{d})=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ Now, $g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$ $1-\frac{2 h}{R}=1-\frac{d}{R}$ $-\frac{2 h}{R}=-\frac{d}{R}$ $h=\frac{d}{2}$ $h=\frac{80}{2}$ $h=40 \mathrm{~km}$
138383
When you move from equator to poles, the value of acceleration due to gravity (g)
1 increases
2 decreases
3 remains the same
4 increases then decreases
Explanation:
A Value of $g$ varies with latitude $(\lambda)$ as $\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2} \cos ^{2} \lambda$ $\therefore$ At poles $\lambda=90^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}$ (maximum) At equator $\lambda=0^{\circ} \Rightarrow \mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R}_{\mathrm{e}} \omega^{2}$ (minimum) Also, the distance between equations is greater than the distance between poles by $21 \mathrm{~km}$.
SRMJEE - 2013
Gravitation
138384
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:
A At ball will execute S.H.M. $\mathrm{T}=2 \pi \sqrt{\frac{\text { Displacement }}{\text { Acceleration }}}$ $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{R}}{\mathrm{g}}}$ $\mathrm{T}=2 \pi \sqrt{\frac{6400 \times 10^{3}}{9.8}} \times \frac{1}{60}$ $\mathrm{~T}=84.63$ So, time taken to reach the other end is $t=\frac{84.63}{2}$ $t=42.31 \mathrm{~min}$
SRMJEE - 2014
Gravitation
138385
The planets with radii $R_{1}$ and $R_{2}$ have densities $\rho_{1}, \rho_{2}$ respectively. Their atmospheric pressure are $p_{1}$ and $p_{2}$ respectively. Therefore the ratio of masses of their atmospheres, neglecting variation of $g$ within the limits of atmosphere is
D We know that, Acceleration due to gravity- $\mathrm{g}=\frac{\mathrm{Gm}}{\mathrm{R}^{2}}$ $\mathrm{~g}=\frac{4}{3} \pi \rho \mathrm{GR}$ $\therefore \quad \frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\rho_{1} \mathrm{R}_{1}}{\rho_{2} \mathrm{R}_{2}}$ Pressure $=\frac{\mathrm{w}}{\mathrm{S}} \Rightarrow \mathrm{p}=\frac{\mathrm{mg}}{\mathrm{S}} \Rightarrow \mathrm{m}=\frac{\mathrm{pS}}{\mathrm{g}}$ Where $\mathrm{w}=$ weight of atmosphere. $\mathrm{S}=$ Surface area of the planet. $\frac{m_{1}}{m_{2}}=\frac{p_{1} S_{1} g_{2}}{p_{2} S_{2} g_{1}}$ $\frac{m_{1}}{m_{2}}=\frac{p_{1} \cdot\left(4 \pi R_{1}^{2}\right)}{p_{2}\left(4 \pi R_{2}^{2}\right)} \times \frac{\rho_{2} R_{2}}{\rho_{1} R_{1}}$\$\$ $\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{\mathrm{p}_{1} \mathrm{R}_{1} \rho_{2}}{\mathrm{p}_{2} \mathrm{R}_{2} \rho_{1}}$
JIPMER-2017
Gravitation
138386
At what height above the earth's surface, the value of $g$ is same as in a mine $80 \mathrm{~km}$ deep?
1 $20 \mathrm{~km}$
2 $30 \mathrm{~km}$
3 $40 \mathrm{~km}$
4 $50 \mathrm{~km}$
Explanation:
C We know that, Variation of $\mathrm{g}$ with height $(\mathrm{h})$, $\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ Variation of $\mathrm{g}$ with depth $(\mathrm{d})=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ Now, $g\left(1-\frac{2 h}{R}\right)=g\left(1-\frac{d}{R}\right)$ $1-\frac{2 h}{R}=1-\frac{d}{R}$ $-\frac{2 h}{R}=-\frac{d}{R}$ $h=\frac{d}{2}$ $h=\frac{80}{2}$ $h=40 \mathrm{~km}$
