138340
If radius of the earth $6347 \mathrm{~km}$, then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface?
1 0.3400
2 0.0340
3 0.0034
4 0.24
Explanation:
B We know that, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ Acceleration due to free fall, $g_{f}=\frac{G_{e}}{R^{2}}-\omega^{2} R$ $g-g_{f}=\frac{G M_{e}}{R^{2}}-\left(\frac{G M_{e}}{R^{2}}-\omega^{2} R\right)$ $g-g_{f}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R$ $g-g_{f}=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}=0.0340$
AIIMS-25.05.2019(M) Shift-1
Gravitation
138341
A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is
1 $1: 4$
2 $1: 5$
3 $1: 2$
4 $3: 2$
Explanation:
C Given, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}$ $\mathrm{G}_{\mathrm{P}}=2 \mathrm{G}_{\mathrm{e}}$ Acceleration due to gravity of the earth having mass $\left(\mathrm{M}_{\mathrm{e}}\right)$ is given by $g_{e}=\frac{G_{e} M_{e}}{R_{e}^{2}}$ $g_{e}=\frac{G_{e}}{R_{e}^{2}} \times \frac{4}{3} \pi R_{e}^{3} \rho_{e}$ $g_{e}=\frac{4}{3} \pi G_{e} R_{e} \rho_{e}$ Acceleration due to planet, $g_{P}=\frac{G_{P} M_{P}}{R_{P}^{2}}$ $g_{P}=\frac{G_{P}}{R_{P}^{2}} \times \frac{4}{3} \pi R_{P}^{3} \rho_{P}$ \(\mathrm{g}_{\mathrm{p}}=\frac{4}{3} \pi \times 2 \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}} \quad\left\{\begin{array}{r}\because \mathrm{G}_{\mathrm{p}}=2 \mathrm{G}_{\mathrm{e}} \\ \rho_{\mathrm{p}}=\rho_{\mathrm{e}}\end{array}\right\}\) $\mathrm{g}_{\mathrm{p}}=\frac{8}{3} \pi \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}}$ From eq.(i) and (ii), $\mathrm{R}_{\mathrm{e}}: \mathrm{R}_{\mathrm{p}}=1: 2$
AIIMS-25.05.2019(E)]**# Shift-2
Gravitation
138342
The acceleration due to gravity on the planet is 9 times the acceleration due to gravity on planet $B$. A man jumps to a height of $2 \mathrm{~m}$ on the surface of $A$. What is the height of jump by the same on the planet $B$ ?
1 $6 \mathrm{~m}$
2 $2 / 3 \mathrm{~m}$
3 $2 / 9 \mathrm{~m}$
4 $18 \mathrm{~m}$
Explanation:
D Given, $\mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}$ The third equation of motion is $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $\mathrm{v}^{2}=2 \mathrm{gh} \quad\{\because \mathrm{u}=0\}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}$ Let's the height of jump on planet $\mathrm{A}$ is $\mathrm{h}_{\mathrm{A}}$ the height of the jump on planet $B$ is $h_{B}$, $\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\mathrm{h}_{\mathrm{B}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{B}}}$ On dividing equations (ii) by equation (iii) $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{2 \mathrm{~g}_{\mathrm{B}}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{\mathrm{g}_{\mathrm{B}}}{9 \mathrm{~g}_{\mathrm{B}}} \quad\left[\because \mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}\right]$ $\mathrm{h}_{\mathrm{B}} =9 \mathrm{~h}_{\mathrm{A}}$ $\mathrm{h}_{\mathrm{B}} =9 \times 2$ $\mathrm{~h}_{\mathrm{B}} =18 \mathrm{~m} \quad\left\{\because \mathrm{h}_{\mathrm{A}}=2 \mathrm{~m}\right\}$
BCECE-2015
Gravitation
138343
The mass of a planet is double and its radius is half compared to that of earth. Assuming $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$ on earth, the acceleration due to gravity at the planet will be-
1 $10 \mathrm{~m} / \mathrm{s}^{2}$
2 $20 \mathrm{~m} / \mathrm{s}^{2}$
3 $40 \mathrm{~m} / \mathrm{s}^{2}$
4 None of these
Explanation:
D Given, $M_{P}=2 M_{e}, R_{P}=\frac{R_{e}}{2}$ As we know acceleration due to gravity on earth surface $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on planet $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Where $\mathrm{Mp}$ is mass of the planet and $\mathrm{Rp}$ is radius of planet On dividing eq $\mathrm{q}^{\mathrm{n}}$ (ii) by (i) we get $\frac{g_{p}}{g_{e}} =\frac{M_{p}}{M_{e}} \times \frac{R_{e}^{2}}{R_{p}^{2}}$ $=\frac{2 M_{e}}{M_{e}} \times \frac{R_{e}^{2}}{\left(\frac{R_{e}}{2}\right)^{2}} \quad\left\{\because M_{p}=2 M_{e}\right\}$ $\frac{g_{p}}{g_{e}} =2 \times(2)^{2}$ $\frac{g_{P}}{g_{e}} =8$ $g_{P} =8 g_{e}$ $g_{P} =8 \times 10=80 \mathrm{~m} / \mathrm{s}^{2} \quad\left\{\because g_{e}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$
BCECE-2013
Gravitation
138344
Find ratio of acceleration due to gravity $g$ at depth $d$ and at height $h$, where $d=2 h$.
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
A If $g_{h}$ is the acceleration due to gravity at a height $h$ about the surface of earth and $g$ be acceleration due to gravity on earths surface, Then, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ It $g_{d}$ be the acceleration due to gravity at a point at depth $d$ below the surface of earth $g_{d}=g\left(1-\frac{d}{R}\right) \quad \quad[\therefore d=2 . h]$ $g_{d}=g\left(1-\frac{2 h}{R}\right)$ Dividing equation (ii) by (i) we get \(\frac{g_d}{g_h}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1\) $\therefore \quad g_{d}: g_{h}=1: 1$
138340
If radius of the earth $6347 \mathrm{~km}$, then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface?
1 0.3400
2 0.0340
3 0.0034
4 0.24
Explanation:
B We know that, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ Acceleration due to free fall, $g_{f}=\frac{G_{e}}{R^{2}}-\omega^{2} R$ $g-g_{f}=\frac{G M_{e}}{R^{2}}-\left(\frac{G M_{e}}{R^{2}}-\omega^{2} R\right)$ $g-g_{f}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R$ $g-g_{f}=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}=0.0340$
AIIMS-25.05.2019(M) Shift-1
Gravitation
138341
A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is
1 $1: 4$
2 $1: 5$
3 $1: 2$
4 $3: 2$
Explanation:
C Given, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}$ $\mathrm{G}_{\mathrm{P}}=2 \mathrm{G}_{\mathrm{e}}$ Acceleration due to gravity of the earth having mass $\left(\mathrm{M}_{\mathrm{e}}\right)$ is given by $g_{e}=\frac{G_{e} M_{e}}{R_{e}^{2}}$ $g_{e}=\frac{G_{e}}{R_{e}^{2}} \times \frac{4}{3} \pi R_{e}^{3} \rho_{e}$ $g_{e}=\frac{4}{3} \pi G_{e} R_{e} \rho_{e}$ Acceleration due to planet, $g_{P}=\frac{G_{P} M_{P}}{R_{P}^{2}}$ $g_{P}=\frac{G_{P}}{R_{P}^{2}} \times \frac{4}{3} \pi R_{P}^{3} \rho_{P}$ \(\mathrm{g}_{\mathrm{p}}=\frac{4}{3} \pi \times 2 \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}} \quad\left\{\begin{array}{r}\because \mathrm{G}_{\mathrm{p}}=2 \mathrm{G}_{\mathrm{e}} \\ \rho_{\mathrm{p}}=\rho_{\mathrm{e}}\end{array}\right\}\) $\mathrm{g}_{\mathrm{p}}=\frac{8}{3} \pi \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}}$ From eq.(i) and (ii), $\mathrm{R}_{\mathrm{e}}: \mathrm{R}_{\mathrm{p}}=1: 2$
AIIMS-25.05.2019(E)]**# Shift-2
Gravitation
138342
The acceleration due to gravity on the planet is 9 times the acceleration due to gravity on planet $B$. A man jumps to a height of $2 \mathrm{~m}$ on the surface of $A$. What is the height of jump by the same on the planet $B$ ?
1 $6 \mathrm{~m}$
2 $2 / 3 \mathrm{~m}$
3 $2 / 9 \mathrm{~m}$
4 $18 \mathrm{~m}$
Explanation:
D Given, $\mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}$ The third equation of motion is $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $\mathrm{v}^{2}=2 \mathrm{gh} \quad\{\because \mathrm{u}=0\}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}$ Let's the height of jump on planet $\mathrm{A}$ is $\mathrm{h}_{\mathrm{A}}$ the height of the jump on planet $B$ is $h_{B}$, $\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\mathrm{h}_{\mathrm{B}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{B}}}$ On dividing equations (ii) by equation (iii) $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{2 \mathrm{~g}_{\mathrm{B}}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{\mathrm{g}_{\mathrm{B}}}{9 \mathrm{~g}_{\mathrm{B}}} \quad\left[\because \mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}\right]$ $\mathrm{h}_{\mathrm{B}} =9 \mathrm{~h}_{\mathrm{A}}$ $\mathrm{h}_{\mathrm{B}} =9 \times 2$ $\mathrm{~h}_{\mathrm{B}} =18 \mathrm{~m} \quad\left\{\because \mathrm{h}_{\mathrm{A}}=2 \mathrm{~m}\right\}$
BCECE-2015
Gravitation
138343
The mass of a planet is double and its radius is half compared to that of earth. Assuming $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$ on earth, the acceleration due to gravity at the planet will be-
1 $10 \mathrm{~m} / \mathrm{s}^{2}$
2 $20 \mathrm{~m} / \mathrm{s}^{2}$
3 $40 \mathrm{~m} / \mathrm{s}^{2}$
4 None of these
Explanation:
D Given, $M_{P}=2 M_{e}, R_{P}=\frac{R_{e}}{2}$ As we know acceleration due to gravity on earth surface $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on planet $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Where $\mathrm{Mp}$ is mass of the planet and $\mathrm{Rp}$ is radius of planet On dividing eq $\mathrm{q}^{\mathrm{n}}$ (ii) by (i) we get $\frac{g_{p}}{g_{e}} =\frac{M_{p}}{M_{e}} \times \frac{R_{e}^{2}}{R_{p}^{2}}$ $=\frac{2 M_{e}}{M_{e}} \times \frac{R_{e}^{2}}{\left(\frac{R_{e}}{2}\right)^{2}} \quad\left\{\because M_{p}=2 M_{e}\right\}$ $\frac{g_{p}}{g_{e}} =2 \times(2)^{2}$ $\frac{g_{P}}{g_{e}} =8$ $g_{P} =8 g_{e}$ $g_{P} =8 \times 10=80 \mathrm{~m} / \mathrm{s}^{2} \quad\left\{\because g_{e}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$
BCECE-2013
Gravitation
138344
Find ratio of acceleration due to gravity $g$ at depth $d$ and at height $h$, where $d=2 h$.
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
A If $g_{h}$ is the acceleration due to gravity at a height $h$ about the surface of earth and $g$ be acceleration due to gravity on earths surface, Then, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ It $g_{d}$ be the acceleration due to gravity at a point at depth $d$ below the surface of earth $g_{d}=g\left(1-\frac{d}{R}\right) \quad \quad[\therefore d=2 . h]$ $g_{d}=g\left(1-\frac{2 h}{R}\right)$ Dividing equation (ii) by (i) we get \(\frac{g_d}{g_h}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1\) $\therefore \quad g_{d}: g_{h}=1: 1$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Gravitation
138340
If radius of the earth $6347 \mathrm{~km}$, then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface?
1 0.3400
2 0.0340
3 0.0034
4 0.24
Explanation:
B We know that, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ Acceleration due to free fall, $g_{f}=\frac{G_{e}}{R^{2}}-\omega^{2} R$ $g-g_{f}=\frac{G M_{e}}{R^{2}}-\left(\frac{G M_{e}}{R^{2}}-\omega^{2} R\right)$ $g-g_{f}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R$ $g-g_{f}=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}=0.0340$
AIIMS-25.05.2019(M) Shift-1
Gravitation
138341
A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is
1 $1: 4$
2 $1: 5$
3 $1: 2$
4 $3: 2$
Explanation:
C Given, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}$ $\mathrm{G}_{\mathrm{P}}=2 \mathrm{G}_{\mathrm{e}}$ Acceleration due to gravity of the earth having mass $\left(\mathrm{M}_{\mathrm{e}}\right)$ is given by $g_{e}=\frac{G_{e} M_{e}}{R_{e}^{2}}$ $g_{e}=\frac{G_{e}}{R_{e}^{2}} \times \frac{4}{3} \pi R_{e}^{3} \rho_{e}$ $g_{e}=\frac{4}{3} \pi G_{e} R_{e} \rho_{e}$ Acceleration due to planet, $g_{P}=\frac{G_{P} M_{P}}{R_{P}^{2}}$ $g_{P}=\frac{G_{P}}{R_{P}^{2}} \times \frac{4}{3} \pi R_{P}^{3} \rho_{P}$ \(\mathrm{g}_{\mathrm{p}}=\frac{4}{3} \pi \times 2 \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}} \quad\left\{\begin{array}{r}\because \mathrm{G}_{\mathrm{p}}=2 \mathrm{G}_{\mathrm{e}} \\ \rho_{\mathrm{p}}=\rho_{\mathrm{e}}\end{array}\right\}\) $\mathrm{g}_{\mathrm{p}}=\frac{8}{3} \pi \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}}$ From eq.(i) and (ii), $\mathrm{R}_{\mathrm{e}}: \mathrm{R}_{\mathrm{p}}=1: 2$
AIIMS-25.05.2019(E)]**# Shift-2
Gravitation
138342
The acceleration due to gravity on the planet is 9 times the acceleration due to gravity on planet $B$. A man jumps to a height of $2 \mathrm{~m}$ on the surface of $A$. What is the height of jump by the same on the planet $B$ ?
1 $6 \mathrm{~m}$
2 $2 / 3 \mathrm{~m}$
3 $2 / 9 \mathrm{~m}$
4 $18 \mathrm{~m}$
Explanation:
D Given, $\mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}$ The third equation of motion is $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $\mathrm{v}^{2}=2 \mathrm{gh} \quad\{\because \mathrm{u}=0\}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}$ Let's the height of jump on planet $\mathrm{A}$ is $\mathrm{h}_{\mathrm{A}}$ the height of the jump on planet $B$ is $h_{B}$, $\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\mathrm{h}_{\mathrm{B}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{B}}}$ On dividing equations (ii) by equation (iii) $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{2 \mathrm{~g}_{\mathrm{B}}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{\mathrm{g}_{\mathrm{B}}}{9 \mathrm{~g}_{\mathrm{B}}} \quad\left[\because \mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}\right]$ $\mathrm{h}_{\mathrm{B}} =9 \mathrm{~h}_{\mathrm{A}}$ $\mathrm{h}_{\mathrm{B}} =9 \times 2$ $\mathrm{~h}_{\mathrm{B}} =18 \mathrm{~m} \quad\left\{\because \mathrm{h}_{\mathrm{A}}=2 \mathrm{~m}\right\}$
BCECE-2015
Gravitation
138343
The mass of a planet is double and its radius is half compared to that of earth. Assuming $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$ on earth, the acceleration due to gravity at the planet will be-
1 $10 \mathrm{~m} / \mathrm{s}^{2}$
2 $20 \mathrm{~m} / \mathrm{s}^{2}$
3 $40 \mathrm{~m} / \mathrm{s}^{2}$
4 None of these
Explanation:
D Given, $M_{P}=2 M_{e}, R_{P}=\frac{R_{e}}{2}$ As we know acceleration due to gravity on earth surface $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on planet $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Where $\mathrm{Mp}$ is mass of the planet and $\mathrm{Rp}$ is radius of planet On dividing eq $\mathrm{q}^{\mathrm{n}}$ (ii) by (i) we get $\frac{g_{p}}{g_{e}} =\frac{M_{p}}{M_{e}} \times \frac{R_{e}^{2}}{R_{p}^{2}}$ $=\frac{2 M_{e}}{M_{e}} \times \frac{R_{e}^{2}}{\left(\frac{R_{e}}{2}\right)^{2}} \quad\left\{\because M_{p}=2 M_{e}\right\}$ $\frac{g_{p}}{g_{e}} =2 \times(2)^{2}$ $\frac{g_{P}}{g_{e}} =8$ $g_{P} =8 g_{e}$ $g_{P} =8 \times 10=80 \mathrm{~m} / \mathrm{s}^{2} \quad\left\{\because g_{e}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$
BCECE-2013
Gravitation
138344
Find ratio of acceleration due to gravity $g$ at depth $d$ and at height $h$, where $d=2 h$.
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
A If $g_{h}$ is the acceleration due to gravity at a height $h$ about the surface of earth and $g$ be acceleration due to gravity on earths surface, Then, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ It $g_{d}$ be the acceleration due to gravity at a point at depth $d$ below the surface of earth $g_{d}=g\left(1-\frac{d}{R}\right) \quad \quad[\therefore d=2 . h]$ $g_{d}=g\left(1-\frac{2 h}{R}\right)$ Dividing equation (ii) by (i) we get \(\frac{g_d}{g_h}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1\) $\therefore \quad g_{d}: g_{h}=1: 1$
138340
If radius of the earth $6347 \mathrm{~km}$, then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface?
1 0.3400
2 0.0340
3 0.0034
4 0.24
Explanation:
B We know that, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ Acceleration due to free fall, $g_{f}=\frac{G_{e}}{R^{2}}-\omega^{2} R$ $g-g_{f}=\frac{G M_{e}}{R^{2}}-\left(\frac{G M_{e}}{R^{2}}-\omega^{2} R\right)$ $g-g_{f}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R$ $g-g_{f}=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}=0.0340$
AIIMS-25.05.2019(M) Shift-1
Gravitation
138341
A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is
1 $1: 4$
2 $1: 5$
3 $1: 2$
4 $3: 2$
Explanation:
C Given, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}$ $\mathrm{G}_{\mathrm{P}}=2 \mathrm{G}_{\mathrm{e}}$ Acceleration due to gravity of the earth having mass $\left(\mathrm{M}_{\mathrm{e}}\right)$ is given by $g_{e}=\frac{G_{e} M_{e}}{R_{e}^{2}}$ $g_{e}=\frac{G_{e}}{R_{e}^{2}} \times \frac{4}{3} \pi R_{e}^{3} \rho_{e}$ $g_{e}=\frac{4}{3} \pi G_{e} R_{e} \rho_{e}$ Acceleration due to planet, $g_{P}=\frac{G_{P} M_{P}}{R_{P}^{2}}$ $g_{P}=\frac{G_{P}}{R_{P}^{2}} \times \frac{4}{3} \pi R_{P}^{3} \rho_{P}$ \(\mathrm{g}_{\mathrm{p}}=\frac{4}{3} \pi \times 2 \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}} \quad\left\{\begin{array}{r}\because \mathrm{G}_{\mathrm{p}}=2 \mathrm{G}_{\mathrm{e}} \\ \rho_{\mathrm{p}}=\rho_{\mathrm{e}}\end{array}\right\}\) $\mathrm{g}_{\mathrm{p}}=\frac{8}{3} \pi \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}}$ From eq.(i) and (ii), $\mathrm{R}_{\mathrm{e}}: \mathrm{R}_{\mathrm{p}}=1: 2$
AIIMS-25.05.2019(E)]**# Shift-2
Gravitation
138342
The acceleration due to gravity on the planet is 9 times the acceleration due to gravity on planet $B$. A man jumps to a height of $2 \mathrm{~m}$ on the surface of $A$. What is the height of jump by the same on the planet $B$ ?
1 $6 \mathrm{~m}$
2 $2 / 3 \mathrm{~m}$
3 $2 / 9 \mathrm{~m}$
4 $18 \mathrm{~m}$
Explanation:
D Given, $\mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}$ The third equation of motion is $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $\mathrm{v}^{2}=2 \mathrm{gh} \quad\{\because \mathrm{u}=0\}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}$ Let's the height of jump on planet $\mathrm{A}$ is $\mathrm{h}_{\mathrm{A}}$ the height of the jump on planet $B$ is $h_{B}$, $\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\mathrm{h}_{\mathrm{B}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{B}}}$ On dividing equations (ii) by equation (iii) $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{2 \mathrm{~g}_{\mathrm{B}}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{\mathrm{g}_{\mathrm{B}}}{9 \mathrm{~g}_{\mathrm{B}}} \quad\left[\because \mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}\right]$ $\mathrm{h}_{\mathrm{B}} =9 \mathrm{~h}_{\mathrm{A}}$ $\mathrm{h}_{\mathrm{B}} =9 \times 2$ $\mathrm{~h}_{\mathrm{B}} =18 \mathrm{~m} \quad\left\{\because \mathrm{h}_{\mathrm{A}}=2 \mathrm{~m}\right\}$
BCECE-2015
Gravitation
138343
The mass of a planet is double and its radius is half compared to that of earth. Assuming $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$ on earth, the acceleration due to gravity at the planet will be-
1 $10 \mathrm{~m} / \mathrm{s}^{2}$
2 $20 \mathrm{~m} / \mathrm{s}^{2}$
3 $40 \mathrm{~m} / \mathrm{s}^{2}$
4 None of these
Explanation:
D Given, $M_{P}=2 M_{e}, R_{P}=\frac{R_{e}}{2}$ As we know acceleration due to gravity on earth surface $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on planet $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Where $\mathrm{Mp}$ is mass of the planet and $\mathrm{Rp}$ is radius of planet On dividing eq $\mathrm{q}^{\mathrm{n}}$ (ii) by (i) we get $\frac{g_{p}}{g_{e}} =\frac{M_{p}}{M_{e}} \times \frac{R_{e}^{2}}{R_{p}^{2}}$ $=\frac{2 M_{e}}{M_{e}} \times \frac{R_{e}^{2}}{\left(\frac{R_{e}}{2}\right)^{2}} \quad\left\{\because M_{p}=2 M_{e}\right\}$ $\frac{g_{p}}{g_{e}} =2 \times(2)^{2}$ $\frac{g_{P}}{g_{e}} =8$ $g_{P} =8 g_{e}$ $g_{P} =8 \times 10=80 \mathrm{~m} / \mathrm{s}^{2} \quad\left\{\because g_{e}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$
BCECE-2013
Gravitation
138344
Find ratio of acceleration due to gravity $g$ at depth $d$ and at height $h$, where $d=2 h$.
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
A If $g_{h}$ is the acceleration due to gravity at a height $h$ about the surface of earth and $g$ be acceleration due to gravity on earths surface, Then, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ It $g_{d}$ be the acceleration due to gravity at a point at depth $d$ below the surface of earth $g_{d}=g\left(1-\frac{d}{R}\right) \quad \quad[\therefore d=2 . h]$ $g_{d}=g\left(1-\frac{2 h}{R}\right)$ Dividing equation (ii) by (i) we get \(\frac{g_d}{g_h}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1\) $\therefore \quad g_{d}: g_{h}=1: 1$
138340
If radius of the earth $6347 \mathrm{~km}$, then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface?
1 0.3400
2 0.0340
3 0.0034
4 0.24
Explanation:
B We know that, Acceleration due to gravity, $\mathrm{g}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^{2}}$ Acceleration due to free fall, $g_{f}=\frac{G_{e}}{R^{2}}-\omega^{2} R$ $g-g_{f}=\frac{G M_{e}}{R^{2}}-\left(\frac{G M_{e}}{R^{2}}-\omega^{2} R\right)$ $g-g_{f}=\omega^{2} R=\left(\frac{2 \pi}{T}\right)^{2} R$ $g-g_{f}=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}=0.0340$
AIIMS-25.05.2019(M) Shift-1
Gravitation
138341
A planet has same density as that of earth and universal gravitational constant $G$ is twice that of earth, the ratio of acceleration due to gravity, is
1 $1: 4$
2 $1: 5$
3 $1: 2$
4 $3: 2$
Explanation:
C Given, $\rho_{\mathrm{P}}=\rho_{\mathrm{e}}$ $\mathrm{G}_{\mathrm{P}}=2 \mathrm{G}_{\mathrm{e}}$ Acceleration due to gravity of the earth having mass $\left(\mathrm{M}_{\mathrm{e}}\right)$ is given by $g_{e}=\frac{G_{e} M_{e}}{R_{e}^{2}}$ $g_{e}=\frac{G_{e}}{R_{e}^{2}} \times \frac{4}{3} \pi R_{e}^{3} \rho_{e}$ $g_{e}=\frac{4}{3} \pi G_{e} R_{e} \rho_{e}$ Acceleration due to planet, $g_{P}=\frac{G_{P} M_{P}}{R_{P}^{2}}$ $g_{P}=\frac{G_{P}}{R_{P}^{2}} \times \frac{4}{3} \pi R_{P}^{3} \rho_{P}$ \(\mathrm{g}_{\mathrm{p}}=\frac{4}{3} \pi \times 2 \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}} \quad\left\{\begin{array}{r}\because \mathrm{G}_{\mathrm{p}}=2 \mathrm{G}_{\mathrm{e}} \\ \rho_{\mathrm{p}}=\rho_{\mathrm{e}}\end{array}\right\}\) $\mathrm{g}_{\mathrm{p}}=\frac{8}{3} \pi \mathrm{G}_{\mathrm{e}} \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{e}}$ From eq.(i) and (ii), $\mathrm{R}_{\mathrm{e}}: \mathrm{R}_{\mathrm{p}}=1: 2$
AIIMS-25.05.2019(E)]**# Shift-2
Gravitation
138342
The acceleration due to gravity on the planet is 9 times the acceleration due to gravity on planet $B$. A man jumps to a height of $2 \mathrm{~m}$ on the surface of $A$. What is the height of jump by the same on the planet $B$ ?
1 $6 \mathrm{~m}$
2 $2 / 3 \mathrm{~m}$
3 $2 / 9 \mathrm{~m}$
4 $18 \mathrm{~m}$
Explanation:
D Given, $\mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}$ The third equation of motion is $\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}$ $\mathrm{v}^{2}=2 \mathrm{gh} \quad\{\because \mathrm{u}=0\}$ $\mathrm{h}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}}$ Let's the height of jump on planet $\mathrm{A}$ is $\mathrm{h}_{\mathrm{A}}$ the height of the jump on planet $B$ is $h_{B}$, $\mathrm{h}_{\mathrm{A}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\mathrm{h}_{\mathrm{B}}=\frac{\mathrm{v}^{2}}{2 \mathrm{~g}_{\mathrm{B}}}$ On dividing equations (ii) by equation (iii) $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{2 \mathrm{~g}_{\mathrm{B}}}{2 \mathrm{~g}_{\mathrm{A}}}$ $\frac{\mathrm{h}_{\mathrm{A}}}{\mathrm{h}_{\mathrm{B}}} =\frac{\mathrm{g}_{\mathrm{B}}}{9 \mathrm{~g}_{\mathrm{B}}} \quad\left[\because \mathrm{g}_{\mathrm{A}}=9 \mathrm{~g}_{\mathrm{B}}\right]$ $\mathrm{h}_{\mathrm{B}} =9 \mathrm{~h}_{\mathrm{A}}$ $\mathrm{h}_{\mathrm{B}} =9 \times 2$ $\mathrm{~h}_{\mathrm{B}} =18 \mathrm{~m} \quad\left\{\because \mathrm{h}_{\mathrm{A}}=2 \mathrm{~m}\right\}$
BCECE-2015
Gravitation
138343
The mass of a planet is double and its radius is half compared to that of earth. Assuming $g=$ $10 \mathrm{~m} / \mathrm{s}^{2}$ on earth, the acceleration due to gravity at the planet will be-
1 $10 \mathrm{~m} / \mathrm{s}^{2}$
2 $20 \mathrm{~m} / \mathrm{s}^{2}$
3 $40 \mathrm{~m} / \mathrm{s}^{2}$
4 None of these
Explanation:
D Given, $M_{P}=2 M_{e}, R_{P}=\frac{R_{e}}{2}$ As we know acceleration due to gravity on earth surface $\mathrm{g}_{\mathrm{e}}=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}^{2}}$ Acceleration due to gravity on planet $\mathrm{g}_{\mathrm{P}}=\frac{\mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}^{2}}$ Where $\mathrm{Mp}$ is mass of the planet and $\mathrm{Rp}$ is radius of planet On dividing eq $\mathrm{q}^{\mathrm{n}}$ (ii) by (i) we get $\frac{g_{p}}{g_{e}} =\frac{M_{p}}{M_{e}} \times \frac{R_{e}^{2}}{R_{p}^{2}}$ $=\frac{2 M_{e}}{M_{e}} \times \frac{R_{e}^{2}}{\left(\frac{R_{e}}{2}\right)^{2}} \quad\left\{\because M_{p}=2 M_{e}\right\}$ $\frac{g_{p}}{g_{e}} =2 \times(2)^{2}$ $\frac{g_{P}}{g_{e}} =8$ $g_{P} =8 g_{e}$ $g_{P} =8 \times 10=80 \mathrm{~m} / \mathrm{s}^{2} \quad\left\{\because g_{e}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$
BCECE-2013
Gravitation
138344
Find ratio of acceleration due to gravity $g$ at depth $d$ and at height $h$, where $d=2 h$.
1 $1: 1$
2 $1: 2$
3 $2: 1$
4 $1: 4$
Explanation:
A If $g_{h}$ is the acceleration due to gravity at a height $h$ about the surface of earth and $g$ be acceleration due to gravity on earths surface, Then, $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right)$ It $g_{d}$ be the acceleration due to gravity at a point at depth $d$ below the surface of earth $g_{d}=g\left(1-\frac{d}{R}\right) \quad \quad[\therefore d=2 . h]$ $g_{d}=g\left(1-\frac{2 h}{R}\right)$ Dividing equation (ii) by (i) we get \(\frac{g_d}{g_h}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1\) $\therefore \quad g_{d}: g_{h}=1: 1$