NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138312
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of $g$ on the earth's surface would
1 increase by $0.5 \%$
2 increase by $2 \%$
3 decrease by $0.5 \%$
4 decrease by $2 \%$
Explanation:
B We know, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Differentiate both side with respect to $\mathrm{R}$, $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^{3}}$ $\frac{\mathrm{dgR}}{\mathrm{GM}}=-\frac{2 \mathrm{dR}}{\mathrm{R}}$ $\frac{\mathrm{dg}}{\mathrm{g}}=-2 \frac{\mathrm{dR}}{\mathrm{R}}$ [From equation(i)] Given, $\frac{\mathrm{dR}}{\mathrm{R}}=-1 \%$ $\therefore \quad \frac{\mathrm{dg}}{\mathrm{g}}=(-2)(-1) \%$ Hence, it increases by $2 \%$.
IIT Advance 1981
Gravitation
138313
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400 \mathrm{~km}$.
138315
A planet has twice the values of mass and radius than that of the earth. Acceleration due to gravity on the surface of planet is
1 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
2 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
3 $980 \mathrm{~m} / \mathrm{s}^{2}$
4 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, mass of earth $=\mathrm{M}_{\mathrm{e}}$ radius of earth $=R_{e}$ mass of planet $=2 \mathrm{M}_{\mathrm{e}}$ radius of earth $=2 \mathrm{R}_{\mathrm{e}}$ We know that- $g^{\prime}=\frac{G M}{R^{2}}$ $=\frac{2 M_{e}}{\left(2 R_{e}\right)^{2}} \quad\left\{\begin{array}{l} R=2 R_{e} \\ M=2 M_{e} \end{array}\right\}$ $=\frac{2}{4} \times \frac{G_{e}}{R_{e}^{2}} \quad\left\{g=\frac{G_{e}}{R_{e}^{2}}\right\}$ $=\frac{2}{4} \times g$ $=\frac{2}{4} \times 9.8$ $g^{\prime}=4.9 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2012
Gravitation
138316
At what distance from the centre of the earth the value of acceleration due to gravity $g$ will be half that on the surface $(R=$ radius of earth $)$ ?
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $1.414 \mathrm{R}$
4 $0.414 \mathrm{R}$
Explanation:
C Acceleration due to gravity at height $\mathrm{h}$ from earth's surface- $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g}{2}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{1}{\sqrt{2}}=\frac{R}{R+h}$ $\mathrm{R}+\mathrm{h}=\sqrt{2} \mathrm{R}$ $\mathrm{h}=\mathrm{R}(\sqrt{2}-1)$ $[\sqrt{2}=1.414]$ $\mathrm{h}=\mathrm{R}(1.414-1)$ $\mathrm{h}=0.414 \mathrm{R}$ $\therefore$ Distance from the centre of the earth $=\mathrm{R}+0.414 \mathrm{R}$ $=1.414 \mathrm{R}$
138312
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of $g$ on the earth's surface would
1 increase by $0.5 \%$
2 increase by $2 \%$
3 decrease by $0.5 \%$
4 decrease by $2 \%$
Explanation:
B We know, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Differentiate both side with respect to $\mathrm{R}$, $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^{3}}$ $\frac{\mathrm{dgR}}{\mathrm{GM}}=-\frac{2 \mathrm{dR}}{\mathrm{R}}$ $\frac{\mathrm{dg}}{\mathrm{g}}=-2 \frac{\mathrm{dR}}{\mathrm{R}}$ [From equation(i)] Given, $\frac{\mathrm{dR}}{\mathrm{R}}=-1 \%$ $\therefore \quad \frac{\mathrm{dg}}{\mathrm{g}}=(-2)(-1) \%$ Hence, it increases by $2 \%$.
IIT Advance 1981
Gravitation
138313
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400 \mathrm{~km}$.
138315
A planet has twice the values of mass and radius than that of the earth. Acceleration due to gravity on the surface of planet is
1 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
2 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
3 $980 \mathrm{~m} / \mathrm{s}^{2}$
4 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, mass of earth $=\mathrm{M}_{\mathrm{e}}$ radius of earth $=R_{e}$ mass of planet $=2 \mathrm{M}_{\mathrm{e}}$ radius of earth $=2 \mathrm{R}_{\mathrm{e}}$ We know that- $g^{\prime}=\frac{G M}{R^{2}}$ $=\frac{2 M_{e}}{\left(2 R_{e}\right)^{2}} \quad\left\{\begin{array}{l} R=2 R_{e} \\ M=2 M_{e} \end{array}\right\}$ $=\frac{2}{4} \times \frac{G_{e}}{R_{e}^{2}} \quad\left\{g=\frac{G_{e}}{R_{e}^{2}}\right\}$ $=\frac{2}{4} \times g$ $=\frac{2}{4} \times 9.8$ $g^{\prime}=4.9 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2012
Gravitation
138316
At what distance from the centre of the earth the value of acceleration due to gravity $g$ will be half that on the surface $(R=$ radius of earth $)$ ?
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $1.414 \mathrm{R}$
4 $0.414 \mathrm{R}$
Explanation:
C Acceleration due to gravity at height $\mathrm{h}$ from earth's surface- $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g}{2}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{1}{\sqrt{2}}=\frac{R}{R+h}$ $\mathrm{R}+\mathrm{h}=\sqrt{2} \mathrm{R}$ $\mathrm{h}=\mathrm{R}(\sqrt{2}-1)$ $[\sqrt{2}=1.414]$ $\mathrm{h}=\mathrm{R}(1.414-1)$ $\mathrm{h}=0.414 \mathrm{R}$ $\therefore$ Distance from the centre of the earth $=\mathrm{R}+0.414 \mathrm{R}$ $=1.414 \mathrm{R}$
138312
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of $g$ on the earth's surface would
1 increase by $0.5 \%$
2 increase by $2 \%$
3 decrease by $0.5 \%$
4 decrease by $2 \%$
Explanation:
B We know, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Differentiate both side with respect to $\mathrm{R}$, $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^{3}}$ $\frac{\mathrm{dgR}}{\mathrm{GM}}=-\frac{2 \mathrm{dR}}{\mathrm{R}}$ $\frac{\mathrm{dg}}{\mathrm{g}}=-2 \frac{\mathrm{dR}}{\mathrm{R}}$ [From equation(i)] Given, $\frac{\mathrm{dR}}{\mathrm{R}}=-1 \%$ $\therefore \quad \frac{\mathrm{dg}}{\mathrm{g}}=(-2)(-1) \%$ Hence, it increases by $2 \%$.
IIT Advance 1981
Gravitation
138313
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400 \mathrm{~km}$.
138315
A planet has twice the values of mass and radius than that of the earth. Acceleration due to gravity on the surface of planet is
1 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
2 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
3 $980 \mathrm{~m} / \mathrm{s}^{2}$
4 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, mass of earth $=\mathrm{M}_{\mathrm{e}}$ radius of earth $=R_{e}$ mass of planet $=2 \mathrm{M}_{\mathrm{e}}$ radius of earth $=2 \mathrm{R}_{\mathrm{e}}$ We know that- $g^{\prime}=\frac{G M}{R^{2}}$ $=\frac{2 M_{e}}{\left(2 R_{e}\right)^{2}} \quad\left\{\begin{array}{l} R=2 R_{e} \\ M=2 M_{e} \end{array}\right\}$ $=\frac{2}{4} \times \frac{G_{e}}{R_{e}^{2}} \quad\left\{g=\frac{G_{e}}{R_{e}^{2}}\right\}$ $=\frac{2}{4} \times g$ $=\frac{2}{4} \times 9.8$ $g^{\prime}=4.9 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2012
Gravitation
138316
At what distance from the centre of the earth the value of acceleration due to gravity $g$ will be half that on the surface $(R=$ radius of earth $)$ ?
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $1.414 \mathrm{R}$
4 $0.414 \mathrm{R}$
Explanation:
C Acceleration due to gravity at height $\mathrm{h}$ from earth's surface- $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g}{2}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{1}{\sqrt{2}}=\frac{R}{R+h}$ $\mathrm{R}+\mathrm{h}=\sqrt{2} \mathrm{R}$ $\mathrm{h}=\mathrm{R}(\sqrt{2}-1)$ $[\sqrt{2}=1.414]$ $\mathrm{h}=\mathrm{R}(1.414-1)$ $\mathrm{h}=0.414 \mathrm{R}$ $\therefore$ Distance from the centre of the earth $=\mathrm{R}+0.414 \mathrm{R}$ $=1.414 \mathrm{R}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138312
If the radius of the earth were to shrink by one percent, its mass remaining the same, the value of $g$ on the earth's surface would
1 increase by $0.5 \%$
2 increase by $2 \%$
3 decrease by $0.5 \%$
4 decrease by $2 \%$
Explanation:
B We know, $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$ Differentiate both side with respect to $\mathrm{R}$, $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{\mathrm{d}}{\mathrm{dR}}\left(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\right)$ $\frac{\mathrm{dg}}{\mathrm{dR}}=\frac{-2 \mathrm{GM}}{\mathrm{R}^{3}}$ $\frac{\mathrm{dgR}}{\mathrm{GM}}=-\frac{2 \mathrm{dR}}{\mathrm{R}}$ $\frac{\mathrm{dg}}{\mathrm{g}}=-2 \frac{\mathrm{dR}}{\mathrm{R}}$ [From equation(i)] Given, $\frac{\mathrm{dR}}{\mathrm{R}}=-1 \%$ $\therefore \quad \frac{\mathrm{dg}}{\mathrm{g}}=(-2)(-1) \%$ Hence, it increases by $2 \%$.
IIT Advance 1981
Gravitation
138313
What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\frac{3}{5}$ of the present weight at the equator. Equatorial radius of the earth is $6400 \mathrm{~km}$.
138315
A planet has twice the values of mass and radius than that of the earth. Acceleration due to gravity on the surface of planet is
1 $9.8 \mathrm{~m} / \mathrm{s}^{2}$
2 $4.9 \mathrm{~m} / \mathrm{s}^{2}$
3 $980 \mathrm{~m} / \mathrm{s}^{2}$
4 $19.6 \mathrm{~m} / \mathrm{s}^{2}$
Explanation:
B Given, mass of earth $=\mathrm{M}_{\mathrm{e}}$ radius of earth $=R_{e}$ mass of planet $=2 \mathrm{M}_{\mathrm{e}}$ radius of earth $=2 \mathrm{R}_{\mathrm{e}}$ We know that- $g^{\prime}=\frac{G M}{R^{2}}$ $=\frac{2 M_{e}}{\left(2 R_{e}\right)^{2}} \quad\left\{\begin{array}{l} R=2 R_{e} \\ M=2 M_{e} \end{array}\right\}$ $=\frac{2}{4} \times \frac{G_{e}}{R_{e}^{2}} \quad\left\{g=\frac{G_{e}}{R_{e}^{2}}\right\}$ $=\frac{2}{4} \times g$ $=\frac{2}{4} \times 9.8$ $g^{\prime}=4.9 \mathrm{~m} / \mathrm{s}^{2}$
CG PET- 2012
Gravitation
138316
At what distance from the centre of the earth the value of acceleration due to gravity $g$ will be half that on the surface $(R=$ radius of earth $)$ ?
1 $2 \mathrm{R}$
2 $\mathrm{R}$
3 $1.414 \mathrm{R}$
4 $0.414 \mathrm{R}$
Explanation:
C Acceleration due to gravity at height $\mathrm{h}$ from earth's surface- $g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{g}{2}=g\left(\frac{R}{R+h}\right)^{2}$ $\frac{1}{\sqrt{2}}=\frac{R}{R+h}$ $\mathrm{R}+\mathrm{h}=\sqrt{2} \mathrm{R}$ $\mathrm{h}=\mathrm{R}(\sqrt{2}-1)$ $[\sqrt{2}=1.414]$ $\mathrm{h}=\mathrm{R}(1.414-1)$ $\mathrm{h}=0.414 \mathrm{R}$ $\therefore$ Distance from the centre of the earth $=\mathrm{R}+0.414 \mathrm{R}$ $=1.414 \mathrm{R}$
