138350
If both the mass and radius of the earth decreases by $1 \%$, then the :
1 escape velocity would decrease
2 escape velocity would increase
3 acceleration due to gravity would increase
4 acceleration due to gravity would decrease
Explanation:
C We know, Escape velocity $\left(\mathrm{V}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{Gm}}{\mathrm{R}_{\mathrm{e}}}}$ If both mass and radius decrease by $1 \%$. Then, escape velocity remains unchanged. Consider earth to be a perfect sphere. Then, Acceleration due to gravity at its surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ If the mass and radius of the earth decreased by $1 \%$ $M^{\prime}=M-\frac{M}{100}=\frac{99}{100} M$ $M^{\prime}=0.99 M$ $R^{\prime}=R-\frac{R}{100}=\frac{99}{100} R$ $R^{\prime}=0.99 R$ Now, New acceleration due to gravity $\left(g^{\prime}\right)=\frac{\mathrm{GM}^{\prime}}{(\mathrm{R})^{2}}$ $g^{\prime}=\frac{G(0.99 M)}{(0.99 R)^{2}}$ $g^{\prime}=1.01 \times \frac{G M}{R^{2}}=1.01 g$ Percentage increase in the acceleration due to gravity is given by - $=\frac{g^{\prime}-g}{g} \times 100$ $=\frac{0.01 g}{g} \times 100=1 \%$ Hence, the acceleration due to gravity increases when the mass and radius decreases.
Karnataka CET-2001
Gravitation
138351
A man weighs $60 \mathrm{~kg}$ at earth surface. At what height above the earth's surface weight become $30 \mathrm{~kg}$ ? Given radius of earth is $6400 \mathrm{~km}$.
1 $2624 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $2020 \mathrm{~km}$
4 none of these
Explanation:
D We know, Weight $(W)=m g$ Now, New weight $\left(\mathrm{W}^{\prime}\right)=\mathrm{mg}$ $=\frac{\mathrm{mg}}{2}=30 \mathrm{~kg} \quad\left[\therefore \mathrm{g}^{\prime}=\frac{\mathrm{g}}{2}\right]$ At surface of earth, Acceleration due to gravity $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ At height (h), Acceleration due to gravity $g^{\prime}=\frac{G M}{(R+h)^{2}}$ Dividing equation (ii) with equation (i), $\frac{g^{\prime}}{g} =\frac{G M}{\frac{(R+h)^{2}}{\frac{G M}{R^{2}}}}$ $=\frac{1}{\left(1+\frac{h}{R}\right)^{2}}$ Now, $\frac{\mathrm{g}}{2 \mathrm{~g}} =\frac{1}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}$ $2 =\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $\sqrt{2} =1+\frac{\mathrm{h}}{\mathrm{R}}$ $\frac{\mathrm{h}}{\mathrm{R}}=1.414-1 \quad[\therefore \sqrt{2}=1.414]$ $\frac{\mathrm{h}}{\mathrm{R}}=0.414$ $\mathrm{h}=6400 \times 0.414$ $=2649.6 \mathrm{~km}$
J and K CET- 2002
Gravitation
138353
Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $\omega$, then $g_{p}-g_{e}$ is given by.
1 $\frac{\omega^{2}}{\mathrm{R}}$
2 $\mathrm{R} \omega^{2}$
3 $\mathrm{R}^{2} \omega^{2}$
4 $\frac{\omega^{2}}{\mathrm{R}^{2}}$
Explanation:
B We know, $\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{\mathrm{R} \omega^{2}}{\mathrm{~g}} \cos ^{2} \lambda\right]$ At the equator, $\lambda=0$ $g_{e}=g\left[1-\frac{R \omega^{2}}{g}\right]$ $g_{e}=g-R \omega^{2}$ At the pole, $\lambda=90^{\circ}$ $g_{p}=g\left[1-\frac{R \omega^{2}}{g} \cos ^{2} 90^{\circ}\right]$ $g_{p}=g$ Now, $\quad \mathrm{g}_{\mathrm{p}}-\mathrm{g}_{\mathrm{e}}=\mathrm{g}-\left(\mathrm{g}-\mathrm{R} \omega^{2}\right)$ $=\mathrm{g}-\mathrm{g}+\mathrm{R} \omega^{2}$ $=\mathrm{R} \omega^{2}$
J and K CET- 2000
Gravitation
138354
If the density of earth is doubled keeping its radius constant then acceleration due to gravity will be $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
2 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
3 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.45 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity at earth surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ And density $(\rho)=\frac{\operatorname{Mass}(\mathrm{M})}{\operatorname{Volume}(\mathrm{V})}$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Then, $\quad \mathrm{g}_{1}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho_{1}}{\mathrm{R}^{2}}$ $\mathrm{g}_{1}=\frac{4 \mathrm{G} \pi \rho_{1} \mathrm{R}}{3}$ And $\mathrm{g}_{2}=\frac{4 \mathrm{G} \pi \rho_{2} \mathrm{R}}{3}$ On dividing equation (ii) by (i), we get - $\frac{g_{2}}{g_{1}}=\frac{\rho_{2}}{\rho_{1}}$ It is given that the density of earth gets doubled. Then, $\rho_{2}=2 \rho_{1}$ Now, $\quad \frac{g_{2}}{g_{1}}=\frac{2 \rho_{1}}{\rho_{1}}$ $\frac{\mathrm{g}_{2}}{\mathrm{~g}_{1}}=2$ $\mathrm{~g}_{2}=2 \mathrm{~g}_{1}$ $\mathrm{~g}_{2}=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}^{2}$
138350
If both the mass and radius of the earth decreases by $1 \%$, then the :
1 escape velocity would decrease
2 escape velocity would increase
3 acceleration due to gravity would increase
4 acceleration due to gravity would decrease
Explanation:
C We know, Escape velocity $\left(\mathrm{V}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{Gm}}{\mathrm{R}_{\mathrm{e}}}}$ If both mass and radius decrease by $1 \%$. Then, escape velocity remains unchanged. Consider earth to be a perfect sphere. Then, Acceleration due to gravity at its surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ If the mass and radius of the earth decreased by $1 \%$ $M^{\prime}=M-\frac{M}{100}=\frac{99}{100} M$ $M^{\prime}=0.99 M$ $R^{\prime}=R-\frac{R}{100}=\frac{99}{100} R$ $R^{\prime}=0.99 R$ Now, New acceleration due to gravity $\left(g^{\prime}\right)=\frac{\mathrm{GM}^{\prime}}{(\mathrm{R})^{2}}$ $g^{\prime}=\frac{G(0.99 M)}{(0.99 R)^{2}}$ $g^{\prime}=1.01 \times \frac{G M}{R^{2}}=1.01 g$ Percentage increase in the acceleration due to gravity is given by - $=\frac{g^{\prime}-g}{g} \times 100$ $=\frac{0.01 g}{g} \times 100=1 \%$ Hence, the acceleration due to gravity increases when the mass and radius decreases.
Karnataka CET-2001
Gravitation
138351
A man weighs $60 \mathrm{~kg}$ at earth surface. At what height above the earth's surface weight become $30 \mathrm{~kg}$ ? Given radius of earth is $6400 \mathrm{~km}$.
1 $2624 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $2020 \mathrm{~km}$
4 none of these
Explanation:
D We know, Weight $(W)=m g$ Now, New weight $\left(\mathrm{W}^{\prime}\right)=\mathrm{mg}$ $=\frac{\mathrm{mg}}{2}=30 \mathrm{~kg} \quad\left[\therefore \mathrm{g}^{\prime}=\frac{\mathrm{g}}{2}\right]$ At surface of earth, Acceleration due to gravity $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ At height (h), Acceleration due to gravity $g^{\prime}=\frac{G M}{(R+h)^{2}}$ Dividing equation (ii) with equation (i), $\frac{g^{\prime}}{g} =\frac{G M}{\frac{(R+h)^{2}}{\frac{G M}{R^{2}}}}$ $=\frac{1}{\left(1+\frac{h}{R}\right)^{2}}$ Now, $\frac{\mathrm{g}}{2 \mathrm{~g}} =\frac{1}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}$ $2 =\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $\sqrt{2} =1+\frac{\mathrm{h}}{\mathrm{R}}$ $\frac{\mathrm{h}}{\mathrm{R}}=1.414-1 \quad[\therefore \sqrt{2}=1.414]$ $\frac{\mathrm{h}}{\mathrm{R}}=0.414$ $\mathrm{h}=6400 \times 0.414$ $=2649.6 \mathrm{~km}$
J and K CET- 2002
Gravitation
138353
Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $\omega$, then $g_{p}-g_{e}$ is given by.
1 $\frac{\omega^{2}}{\mathrm{R}}$
2 $\mathrm{R} \omega^{2}$
3 $\mathrm{R}^{2} \omega^{2}$
4 $\frac{\omega^{2}}{\mathrm{R}^{2}}$
Explanation:
B We know, $\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{\mathrm{R} \omega^{2}}{\mathrm{~g}} \cos ^{2} \lambda\right]$ At the equator, $\lambda=0$ $g_{e}=g\left[1-\frac{R \omega^{2}}{g}\right]$ $g_{e}=g-R \omega^{2}$ At the pole, $\lambda=90^{\circ}$ $g_{p}=g\left[1-\frac{R \omega^{2}}{g} \cos ^{2} 90^{\circ}\right]$ $g_{p}=g$ Now, $\quad \mathrm{g}_{\mathrm{p}}-\mathrm{g}_{\mathrm{e}}=\mathrm{g}-\left(\mathrm{g}-\mathrm{R} \omega^{2}\right)$ $=\mathrm{g}-\mathrm{g}+\mathrm{R} \omega^{2}$ $=\mathrm{R} \omega^{2}$
J and K CET- 2000
Gravitation
138354
If the density of earth is doubled keeping its radius constant then acceleration due to gravity will be $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
2 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
3 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.45 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity at earth surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ And density $(\rho)=\frac{\operatorname{Mass}(\mathrm{M})}{\operatorname{Volume}(\mathrm{V})}$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Then, $\quad \mathrm{g}_{1}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho_{1}}{\mathrm{R}^{2}}$ $\mathrm{g}_{1}=\frac{4 \mathrm{G} \pi \rho_{1} \mathrm{R}}{3}$ And $\mathrm{g}_{2}=\frac{4 \mathrm{G} \pi \rho_{2} \mathrm{R}}{3}$ On dividing equation (ii) by (i), we get - $\frac{g_{2}}{g_{1}}=\frac{\rho_{2}}{\rho_{1}}$ It is given that the density of earth gets doubled. Then, $\rho_{2}=2 \rho_{1}$ Now, $\quad \frac{g_{2}}{g_{1}}=\frac{2 \rho_{1}}{\rho_{1}}$ $\frac{\mathrm{g}_{2}}{\mathrm{~g}_{1}}=2$ $\mathrm{~g}_{2}=2 \mathrm{~g}_{1}$ $\mathrm{~g}_{2}=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}^{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138350
If both the mass and radius of the earth decreases by $1 \%$, then the :
1 escape velocity would decrease
2 escape velocity would increase
3 acceleration due to gravity would increase
4 acceleration due to gravity would decrease
Explanation:
C We know, Escape velocity $\left(\mathrm{V}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{Gm}}{\mathrm{R}_{\mathrm{e}}}}$ If both mass and radius decrease by $1 \%$. Then, escape velocity remains unchanged. Consider earth to be a perfect sphere. Then, Acceleration due to gravity at its surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ If the mass and radius of the earth decreased by $1 \%$ $M^{\prime}=M-\frac{M}{100}=\frac{99}{100} M$ $M^{\prime}=0.99 M$ $R^{\prime}=R-\frac{R}{100}=\frac{99}{100} R$ $R^{\prime}=0.99 R$ Now, New acceleration due to gravity $\left(g^{\prime}\right)=\frac{\mathrm{GM}^{\prime}}{(\mathrm{R})^{2}}$ $g^{\prime}=\frac{G(0.99 M)}{(0.99 R)^{2}}$ $g^{\prime}=1.01 \times \frac{G M}{R^{2}}=1.01 g$ Percentage increase in the acceleration due to gravity is given by - $=\frac{g^{\prime}-g}{g} \times 100$ $=\frac{0.01 g}{g} \times 100=1 \%$ Hence, the acceleration due to gravity increases when the mass and radius decreases.
Karnataka CET-2001
Gravitation
138351
A man weighs $60 \mathrm{~kg}$ at earth surface. At what height above the earth's surface weight become $30 \mathrm{~kg}$ ? Given radius of earth is $6400 \mathrm{~km}$.
1 $2624 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $2020 \mathrm{~km}$
4 none of these
Explanation:
D We know, Weight $(W)=m g$ Now, New weight $\left(\mathrm{W}^{\prime}\right)=\mathrm{mg}$ $=\frac{\mathrm{mg}}{2}=30 \mathrm{~kg} \quad\left[\therefore \mathrm{g}^{\prime}=\frac{\mathrm{g}}{2}\right]$ At surface of earth, Acceleration due to gravity $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ At height (h), Acceleration due to gravity $g^{\prime}=\frac{G M}{(R+h)^{2}}$ Dividing equation (ii) with equation (i), $\frac{g^{\prime}}{g} =\frac{G M}{\frac{(R+h)^{2}}{\frac{G M}{R^{2}}}}$ $=\frac{1}{\left(1+\frac{h}{R}\right)^{2}}$ Now, $\frac{\mathrm{g}}{2 \mathrm{~g}} =\frac{1}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}$ $2 =\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $\sqrt{2} =1+\frac{\mathrm{h}}{\mathrm{R}}$ $\frac{\mathrm{h}}{\mathrm{R}}=1.414-1 \quad[\therefore \sqrt{2}=1.414]$ $\frac{\mathrm{h}}{\mathrm{R}}=0.414$ $\mathrm{h}=6400 \times 0.414$ $=2649.6 \mathrm{~km}$
J and K CET- 2002
Gravitation
138353
Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $\omega$, then $g_{p}-g_{e}$ is given by.
1 $\frac{\omega^{2}}{\mathrm{R}}$
2 $\mathrm{R} \omega^{2}$
3 $\mathrm{R}^{2} \omega^{2}$
4 $\frac{\omega^{2}}{\mathrm{R}^{2}}$
Explanation:
B We know, $\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{\mathrm{R} \omega^{2}}{\mathrm{~g}} \cos ^{2} \lambda\right]$ At the equator, $\lambda=0$ $g_{e}=g\left[1-\frac{R \omega^{2}}{g}\right]$ $g_{e}=g-R \omega^{2}$ At the pole, $\lambda=90^{\circ}$ $g_{p}=g\left[1-\frac{R \omega^{2}}{g} \cos ^{2} 90^{\circ}\right]$ $g_{p}=g$ Now, $\quad \mathrm{g}_{\mathrm{p}}-\mathrm{g}_{\mathrm{e}}=\mathrm{g}-\left(\mathrm{g}-\mathrm{R} \omega^{2}\right)$ $=\mathrm{g}-\mathrm{g}+\mathrm{R} \omega^{2}$ $=\mathrm{R} \omega^{2}$
J and K CET- 2000
Gravitation
138354
If the density of earth is doubled keeping its radius constant then acceleration due to gravity will be $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
2 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
3 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.45 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity at earth surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ And density $(\rho)=\frac{\operatorname{Mass}(\mathrm{M})}{\operatorname{Volume}(\mathrm{V})}$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Then, $\quad \mathrm{g}_{1}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho_{1}}{\mathrm{R}^{2}}$ $\mathrm{g}_{1}=\frac{4 \mathrm{G} \pi \rho_{1} \mathrm{R}}{3}$ And $\mathrm{g}_{2}=\frac{4 \mathrm{G} \pi \rho_{2} \mathrm{R}}{3}$ On dividing equation (ii) by (i), we get - $\frac{g_{2}}{g_{1}}=\frac{\rho_{2}}{\rho_{1}}$ It is given that the density of earth gets doubled. Then, $\rho_{2}=2 \rho_{1}$ Now, $\quad \frac{g_{2}}{g_{1}}=\frac{2 \rho_{1}}{\rho_{1}}$ $\frac{\mathrm{g}_{2}}{\mathrm{~g}_{1}}=2$ $\mathrm{~g}_{2}=2 \mathrm{~g}_{1}$ $\mathrm{~g}_{2}=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}^{2}$
138350
If both the mass and radius of the earth decreases by $1 \%$, then the :
1 escape velocity would decrease
2 escape velocity would increase
3 acceleration due to gravity would increase
4 acceleration due to gravity would decrease
Explanation:
C We know, Escape velocity $\left(\mathrm{V}_{\mathrm{e}}\right)=\sqrt{\frac{2 \mathrm{Gm}}{\mathrm{R}_{\mathrm{e}}}}$ If both mass and radius decrease by $1 \%$. Then, escape velocity remains unchanged. Consider earth to be a perfect sphere. Then, Acceleration due to gravity at its surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ If the mass and radius of the earth decreased by $1 \%$ $M^{\prime}=M-\frac{M}{100}=\frac{99}{100} M$ $M^{\prime}=0.99 M$ $R^{\prime}=R-\frac{R}{100}=\frac{99}{100} R$ $R^{\prime}=0.99 R$ Now, New acceleration due to gravity $\left(g^{\prime}\right)=\frac{\mathrm{GM}^{\prime}}{(\mathrm{R})^{2}}$ $g^{\prime}=\frac{G(0.99 M)}{(0.99 R)^{2}}$ $g^{\prime}=1.01 \times \frac{G M}{R^{2}}=1.01 g$ Percentage increase in the acceleration due to gravity is given by - $=\frac{g^{\prime}-g}{g} \times 100$ $=\frac{0.01 g}{g} \times 100=1 \%$ Hence, the acceleration due to gravity increases when the mass and radius decreases.
Karnataka CET-2001
Gravitation
138351
A man weighs $60 \mathrm{~kg}$ at earth surface. At what height above the earth's surface weight become $30 \mathrm{~kg}$ ? Given radius of earth is $6400 \mathrm{~km}$.
1 $2624 \mathrm{~km}$
2 $3000 \mathrm{~km}$
3 $2020 \mathrm{~km}$
4 none of these
Explanation:
D We know, Weight $(W)=m g$ Now, New weight $\left(\mathrm{W}^{\prime}\right)=\mathrm{mg}$ $=\frac{\mathrm{mg}}{2}=30 \mathrm{~kg} \quad\left[\therefore \mathrm{g}^{\prime}=\frac{\mathrm{g}}{2}\right]$ At surface of earth, Acceleration due to gravity $g=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ At height (h), Acceleration due to gravity $g^{\prime}=\frac{G M}{(R+h)^{2}}$ Dividing equation (ii) with equation (i), $\frac{g^{\prime}}{g} =\frac{G M}{\frac{(R+h)^{2}}{\frac{G M}{R^{2}}}}$ $=\frac{1}{\left(1+\frac{h}{R}\right)^{2}}$ Now, $\frac{\mathrm{g}}{2 \mathrm{~g}} =\frac{1}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}$ $2 =\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}$ $\sqrt{2} =1+\frac{\mathrm{h}}{\mathrm{R}}$ $\frac{\mathrm{h}}{\mathrm{R}}=1.414-1 \quad[\therefore \sqrt{2}=1.414]$ $\frac{\mathrm{h}}{\mathrm{R}}=0.414$ $\mathrm{h}=6400 \times 0.414$ $=2649.6 \mathrm{~km}$
J and K CET- 2002
Gravitation
138353
Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $\omega$, then $g_{p}-g_{e}$ is given by.
1 $\frac{\omega^{2}}{\mathrm{R}}$
2 $\mathrm{R} \omega^{2}$
3 $\mathrm{R}^{2} \omega^{2}$
4 $\frac{\omega^{2}}{\mathrm{R}^{2}}$
Explanation:
B We know, $\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{\mathrm{R} \omega^{2}}{\mathrm{~g}} \cos ^{2} \lambda\right]$ At the equator, $\lambda=0$ $g_{e}=g\left[1-\frac{R \omega^{2}}{g}\right]$ $g_{e}=g-R \omega^{2}$ At the pole, $\lambda=90^{\circ}$ $g_{p}=g\left[1-\frac{R \omega^{2}}{g} \cos ^{2} 90^{\circ}\right]$ $g_{p}=g$ Now, $\quad \mathrm{g}_{\mathrm{p}}-\mathrm{g}_{\mathrm{e}}=\mathrm{g}-\left(\mathrm{g}-\mathrm{R} \omega^{2}\right)$ $=\mathrm{g}-\mathrm{g}+\mathrm{R} \omega^{2}$ $=\mathrm{R} \omega^{2}$
J and K CET- 2000
Gravitation
138354
If the density of earth is doubled keeping its radius constant then acceleration due to gravity will be $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
2 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
3 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
4 $2.45 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
A We know, Acceleration due to gravity at earth surface $(\mathrm{g})=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ And density $(\rho)=\frac{\operatorname{Mass}(\mathrm{M})}{\operatorname{Volume}(\mathrm{V})}$ $M=\rho \times \frac{4}{3} \pi R^{3}$ Then, $\quad \mathrm{g}_{1}=\frac{\mathrm{G} \cdot \frac{4}{3} \pi \mathrm{R}^{3} \times \rho_{1}}{\mathrm{R}^{2}}$ $\mathrm{g}_{1}=\frac{4 \mathrm{G} \pi \rho_{1} \mathrm{R}}{3}$ And $\mathrm{g}_{2}=\frac{4 \mathrm{G} \pi \rho_{2} \mathrm{R}}{3}$ On dividing equation (ii) by (i), we get - $\frac{g_{2}}{g_{1}}=\frac{\rho_{2}}{\rho_{1}}$ It is given that the density of earth gets doubled. Then, $\rho_{2}=2 \rho_{1}$ Now, $\quad \frac{g_{2}}{g_{1}}=\frac{2 \rho_{1}}{\rho_{1}}$ $\frac{\mathrm{g}_{2}}{\mathrm{~g}_{1}}=2$ $\mathrm{~g}_{2}=2 \mathrm{~g}_{1}$ $\mathrm{~g}_{2}=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}^{2}$
