138317
The radius of the earth is 4 times that of the moon and its mass is 80 times that of the moon. If the acceleration due to gravity on the surface of the earth is $10 \mathrm{~m} / \mathrm{s}^{2}$, that on the surface of the moon will be
1 $1 \mathrm{~m} / \mathrm{s}^{2}$
2 $2 \mathrm{~m} / \mathrm{s}^{2}$
3 $3 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, Mass of Moon $=\mathrm{M}_{\mathrm{m}}$ Mass of earth $=80 \mathrm{M}_{\mathrm{m}}$ Radius of Moon $=\mathrm{R}_{\mathrm{m}}$ Radius of earth $=4 \mathrm{R}_{\mathrm{m}}$ Gravity on the surface of moon $=g_{m}$ Gravity on the surface of earth $=g_{e}$ We know that, $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}}$ And $\quad \mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{G} \times 80 \times \mathrm{M}_{\mathrm{m}}}{16 \times \mathrm{R}_{\mathrm{m}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{5 \times \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} / \mathrm{R}_{\mathrm{m}}^{2}}{5 \mathrm{G} / \mathrm{R}_{\mathrm{m}}^{2}}=\frac{1}{5}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times \mathrm{g}_{\mathrm{e}}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times 10$ $\mathrm{g}_{\mathrm{m}}=2 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2004
Gravitation
138318
How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
1 $\frac{R}{2}$
2 $\frac{3 R}{4}$
3 $\frac{\mathrm{R}}{4}$
4 $\frac{\mathrm{R}}{3}$
Explanation:
B According to question- If the weight has to be $\frac{1}{4}$ of the original, the new weight is $\frac{\mathrm{mg}}{4}$. At depth $\mathrm{d}, \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{mg}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\frac{\mathrm{mg}}{4}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \quad\left[\because \mathrm{mg}^{\prime}=\frac{\mathrm{mg}}{4}\right]$ $\frac{1}{4}=1-\frac{\mathrm{d}}{\mathrm{R}}$ $\frac{\mathrm{d}}{\mathrm{R}}=1-\frac{1}{4}$ $\frac{\mathrm{d}}{\mathrm{R}}=\frac{3}{4}$ Hence, $\mathrm{d}=\frac{3 \mathrm{R}}{4}$
Manipal UGET -2020
Gravitation
138319
If the density of earth is doubled keeping its radius constant, then acceleration due to gravity $g$ is
1 $20 \mathrm{~m} / \mathrm{s}^{2}$
2 $10 \mathrm{~m} / \mathrm{s}^{2}$
3 $5 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{g}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho}{\mathrm{R}^{2}}$ $\mathrm{~g}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times \rho$ If the density of the earth is doubled keeping its radius is constant then acceleration due to gravity is, $\mathrm{g}^{\prime}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times(2 \rho)$ $\mathrm{g}^{\prime}=2\left(\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{Rd}\right)$ Putting the value from equation (i), we get $\mathrm{g}^{\prime}=2 \mathrm{~g}$ $\mathrm{~g}^{\prime}=2 \times 10$ $\mathrm{~g}^{\prime}=20 \mathrm{~m} / \mathrm{s}^{2}$
Manipal UGET-2019
Gravitation
138320
A body weights $500 \mathrm{~N}$ on the surface of the earth. How much would it weight half way below the surface of the earth?
1 $1000 \mathrm{~N}$
2 $500 \mathrm{~N}$
3 $250 \mathrm{~N}$
4 $125 \mathrm{~N}$
Explanation:
C Given, Weight on surface of earth $(\mathrm{mg})=500 \mathrm{~N}$ Weight below the surface of earth at depth $d=\frac{R}{2}$. Then, Acceleration due to gravity at depth $\left(g^{\prime}\right)=g\left(1-\frac{d}{R}\right)$ $\mathrm{mg}^{\prime}=\mathrm{mg}\left(1-\frac{\left(\frac{\mathrm{R}}{2}\right)}{\mathrm{R}}\right)$ $\mathrm{W}^{\prime}=500\left(1-\frac{1}{2}\right)$ $\mathrm{W}^{\prime}=500 \times \frac{1}{2}$ $\mathrm{~W}^{\prime}=250 \mathrm{~N}$
138317
The radius of the earth is 4 times that of the moon and its mass is 80 times that of the moon. If the acceleration due to gravity on the surface of the earth is $10 \mathrm{~m} / \mathrm{s}^{2}$, that on the surface of the moon will be
1 $1 \mathrm{~m} / \mathrm{s}^{2}$
2 $2 \mathrm{~m} / \mathrm{s}^{2}$
3 $3 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, Mass of Moon $=\mathrm{M}_{\mathrm{m}}$ Mass of earth $=80 \mathrm{M}_{\mathrm{m}}$ Radius of Moon $=\mathrm{R}_{\mathrm{m}}$ Radius of earth $=4 \mathrm{R}_{\mathrm{m}}$ Gravity on the surface of moon $=g_{m}$ Gravity on the surface of earth $=g_{e}$ We know that, $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}}$ And $\quad \mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{G} \times 80 \times \mathrm{M}_{\mathrm{m}}}{16 \times \mathrm{R}_{\mathrm{m}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{5 \times \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} / \mathrm{R}_{\mathrm{m}}^{2}}{5 \mathrm{G} / \mathrm{R}_{\mathrm{m}}^{2}}=\frac{1}{5}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times \mathrm{g}_{\mathrm{e}}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times 10$ $\mathrm{g}_{\mathrm{m}}=2 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2004
Gravitation
138318
How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
1 $\frac{R}{2}$
2 $\frac{3 R}{4}$
3 $\frac{\mathrm{R}}{4}$
4 $\frac{\mathrm{R}}{3}$
Explanation:
B According to question- If the weight has to be $\frac{1}{4}$ of the original, the new weight is $\frac{\mathrm{mg}}{4}$. At depth $\mathrm{d}, \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{mg}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\frac{\mathrm{mg}}{4}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \quad\left[\because \mathrm{mg}^{\prime}=\frac{\mathrm{mg}}{4}\right]$ $\frac{1}{4}=1-\frac{\mathrm{d}}{\mathrm{R}}$ $\frac{\mathrm{d}}{\mathrm{R}}=1-\frac{1}{4}$ $\frac{\mathrm{d}}{\mathrm{R}}=\frac{3}{4}$ Hence, $\mathrm{d}=\frac{3 \mathrm{R}}{4}$
Manipal UGET -2020
Gravitation
138319
If the density of earth is doubled keeping its radius constant, then acceleration due to gravity $g$ is
1 $20 \mathrm{~m} / \mathrm{s}^{2}$
2 $10 \mathrm{~m} / \mathrm{s}^{2}$
3 $5 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{g}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho}{\mathrm{R}^{2}}$ $\mathrm{~g}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times \rho$ If the density of the earth is doubled keeping its radius is constant then acceleration due to gravity is, $\mathrm{g}^{\prime}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times(2 \rho)$ $\mathrm{g}^{\prime}=2\left(\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{Rd}\right)$ Putting the value from equation (i), we get $\mathrm{g}^{\prime}=2 \mathrm{~g}$ $\mathrm{~g}^{\prime}=2 \times 10$ $\mathrm{~g}^{\prime}=20 \mathrm{~m} / \mathrm{s}^{2}$
Manipal UGET-2019
Gravitation
138320
A body weights $500 \mathrm{~N}$ on the surface of the earth. How much would it weight half way below the surface of the earth?
1 $1000 \mathrm{~N}$
2 $500 \mathrm{~N}$
3 $250 \mathrm{~N}$
4 $125 \mathrm{~N}$
Explanation:
C Given, Weight on surface of earth $(\mathrm{mg})=500 \mathrm{~N}$ Weight below the surface of earth at depth $d=\frac{R}{2}$. Then, Acceleration due to gravity at depth $\left(g^{\prime}\right)=g\left(1-\frac{d}{R}\right)$ $\mathrm{mg}^{\prime}=\mathrm{mg}\left(1-\frac{\left(\frac{\mathrm{R}}{2}\right)}{\mathrm{R}}\right)$ $\mathrm{W}^{\prime}=500\left(1-\frac{1}{2}\right)$ $\mathrm{W}^{\prime}=500 \times \frac{1}{2}$ $\mathrm{~W}^{\prime}=250 \mathrm{~N}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138317
The radius of the earth is 4 times that of the moon and its mass is 80 times that of the moon. If the acceleration due to gravity on the surface of the earth is $10 \mathrm{~m} / \mathrm{s}^{2}$, that on the surface of the moon will be
1 $1 \mathrm{~m} / \mathrm{s}^{2}$
2 $2 \mathrm{~m} / \mathrm{s}^{2}$
3 $3 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, Mass of Moon $=\mathrm{M}_{\mathrm{m}}$ Mass of earth $=80 \mathrm{M}_{\mathrm{m}}$ Radius of Moon $=\mathrm{R}_{\mathrm{m}}$ Radius of earth $=4 \mathrm{R}_{\mathrm{m}}$ Gravity on the surface of moon $=g_{m}$ Gravity on the surface of earth $=g_{e}$ We know that, $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}}$ And $\quad \mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{G} \times 80 \times \mathrm{M}_{\mathrm{m}}}{16 \times \mathrm{R}_{\mathrm{m}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{5 \times \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} / \mathrm{R}_{\mathrm{m}}^{2}}{5 \mathrm{G} / \mathrm{R}_{\mathrm{m}}^{2}}=\frac{1}{5}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times \mathrm{g}_{\mathrm{e}}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times 10$ $\mathrm{g}_{\mathrm{m}}=2 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2004
Gravitation
138318
How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
1 $\frac{R}{2}$
2 $\frac{3 R}{4}$
3 $\frac{\mathrm{R}}{4}$
4 $\frac{\mathrm{R}}{3}$
Explanation:
B According to question- If the weight has to be $\frac{1}{4}$ of the original, the new weight is $\frac{\mathrm{mg}}{4}$. At depth $\mathrm{d}, \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{mg}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\frac{\mathrm{mg}}{4}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \quad\left[\because \mathrm{mg}^{\prime}=\frac{\mathrm{mg}}{4}\right]$ $\frac{1}{4}=1-\frac{\mathrm{d}}{\mathrm{R}}$ $\frac{\mathrm{d}}{\mathrm{R}}=1-\frac{1}{4}$ $\frac{\mathrm{d}}{\mathrm{R}}=\frac{3}{4}$ Hence, $\mathrm{d}=\frac{3 \mathrm{R}}{4}$
Manipal UGET -2020
Gravitation
138319
If the density of earth is doubled keeping its radius constant, then acceleration due to gravity $g$ is
1 $20 \mathrm{~m} / \mathrm{s}^{2}$
2 $10 \mathrm{~m} / \mathrm{s}^{2}$
3 $5 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{g}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho}{\mathrm{R}^{2}}$ $\mathrm{~g}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times \rho$ If the density of the earth is doubled keeping its radius is constant then acceleration due to gravity is, $\mathrm{g}^{\prime}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times(2 \rho)$ $\mathrm{g}^{\prime}=2\left(\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{Rd}\right)$ Putting the value from equation (i), we get $\mathrm{g}^{\prime}=2 \mathrm{~g}$ $\mathrm{~g}^{\prime}=2 \times 10$ $\mathrm{~g}^{\prime}=20 \mathrm{~m} / \mathrm{s}^{2}$
Manipal UGET-2019
Gravitation
138320
A body weights $500 \mathrm{~N}$ on the surface of the earth. How much would it weight half way below the surface of the earth?
1 $1000 \mathrm{~N}$
2 $500 \mathrm{~N}$
3 $250 \mathrm{~N}$
4 $125 \mathrm{~N}$
Explanation:
C Given, Weight on surface of earth $(\mathrm{mg})=500 \mathrm{~N}$ Weight below the surface of earth at depth $d=\frac{R}{2}$. Then, Acceleration due to gravity at depth $\left(g^{\prime}\right)=g\left(1-\frac{d}{R}\right)$ $\mathrm{mg}^{\prime}=\mathrm{mg}\left(1-\frac{\left(\frac{\mathrm{R}}{2}\right)}{\mathrm{R}}\right)$ $\mathrm{W}^{\prime}=500\left(1-\frac{1}{2}\right)$ $\mathrm{W}^{\prime}=500 \times \frac{1}{2}$ $\mathrm{~W}^{\prime}=250 \mathrm{~N}$
138317
The radius of the earth is 4 times that of the moon and its mass is 80 times that of the moon. If the acceleration due to gravity on the surface of the earth is $10 \mathrm{~m} / \mathrm{s}^{2}$, that on the surface of the moon will be
1 $1 \mathrm{~m} / \mathrm{s}^{2}$
2 $2 \mathrm{~m} / \mathrm{s}^{2}$
3 $3 \mathrm{~m} / \mathrm{s}^{2}$
4 $4 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, Mass of Moon $=\mathrm{M}_{\mathrm{m}}$ Mass of earth $=80 \mathrm{M}_{\mathrm{m}}$ Radius of Moon $=\mathrm{R}_{\mathrm{m}}$ Radius of earth $=4 \mathrm{R}_{\mathrm{m}}$ Gravity on the surface of moon $=g_{m}$ Gravity on the surface of earth $=g_{e}$ We know that, $\mathrm{g}_{\mathrm{m}}=\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}^{2}}$ And $\quad \mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{G} \times 80 \times \mathrm{M}_{\mathrm{m}}}{16 \times \mathrm{R}_{\mathrm{m}}^{2}}$ $\mathrm{g}_{\mathrm{e}}=\frac{5 \times \mathrm{G} \times \mathrm{M}_{\mathrm{m}}}{\mathrm{R}_{\mathrm{m}}{ }^{2}}$ On dividing equation (i) and (ii), we get- $\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\mathrm{GM}_{\mathrm{m}} / \mathrm{R}_{\mathrm{m}}^{2}}{5 \mathrm{G} / \mathrm{R}_{\mathrm{m}}^{2}}=\frac{1}{5}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times \mathrm{g}_{\mathrm{e}}$ $\mathrm{g}_{\mathrm{m}}=\frac{1}{5} \times 10$ $\mathrm{g}_{\mathrm{m}}=2 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2004
Gravitation
138318
How much deep inside the earth (radius $R$ ) should a man go, so that his weight becomes one-fourth of that on the earth's surface?
1 $\frac{R}{2}$
2 $\frac{3 R}{4}$
3 $\frac{\mathrm{R}}{4}$
4 $\frac{\mathrm{R}}{3}$
Explanation:
B According to question- If the weight has to be $\frac{1}{4}$ of the original, the new weight is $\frac{\mathrm{mg}}{4}$. At depth $\mathrm{d}, \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\mathrm{mg}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)$ $\frac{\mathrm{mg}}{4}=\mathrm{mg}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right) \quad\left[\because \mathrm{mg}^{\prime}=\frac{\mathrm{mg}}{4}\right]$ $\frac{1}{4}=1-\frac{\mathrm{d}}{\mathrm{R}}$ $\frac{\mathrm{d}}{\mathrm{R}}=1-\frac{1}{4}$ $\frac{\mathrm{d}}{\mathrm{R}}=\frac{3}{4}$ Hence, $\mathrm{d}=\frac{3 \mathrm{R}}{4}$
Manipal UGET -2020
Gravitation
138319
If the density of earth is doubled keeping its radius constant, then acceleration due to gravity $g$ is
1 $20 \mathrm{~m} / \mathrm{s}^{2}$
2 $10 \mathrm{~m} / \mathrm{s}^{2}$
3 $5 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.5 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ $\mathrm{g}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho}{\mathrm{R}^{2}}$ $\mathrm{~g}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times \rho$ If the density of the earth is doubled keeping its radius is constant then acceleration due to gravity is, $\mathrm{g}^{\prime}=\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R} \times(2 \rho)$ $\mathrm{g}^{\prime}=2\left(\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{Rd}\right)$ Putting the value from equation (i), we get $\mathrm{g}^{\prime}=2 \mathrm{~g}$ $\mathrm{~g}^{\prime}=2 \times 10$ $\mathrm{~g}^{\prime}=20 \mathrm{~m} / \mathrm{s}^{2}$
Manipal UGET-2019
Gravitation
138320
A body weights $500 \mathrm{~N}$ on the surface of the earth. How much would it weight half way below the surface of the earth?
1 $1000 \mathrm{~N}$
2 $500 \mathrm{~N}$
3 $250 \mathrm{~N}$
4 $125 \mathrm{~N}$
Explanation:
C Given, Weight on surface of earth $(\mathrm{mg})=500 \mathrm{~N}$ Weight below the surface of earth at depth $d=\frac{R}{2}$. Then, Acceleration due to gravity at depth $\left(g^{\prime}\right)=g\left(1-\frac{d}{R}\right)$ $\mathrm{mg}^{\prime}=\mathrm{mg}\left(1-\frac{\left(\frac{\mathrm{R}}{2}\right)}{\mathrm{R}}\right)$ $\mathrm{W}^{\prime}=500\left(1-\frac{1}{2}\right)$ $\mathrm{W}^{\prime}=500 \times \frac{1}{2}$ $\mathrm{~W}^{\prime}=250 \mathrm{~N}$