138464
The magnitudes of gravitational field at distances $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ and $F_{1}$ and $F_{2}$ respectively. Then,
1 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ if $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$
2 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}$ if $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$
3 Both (a) and (b)
4 None of the above
Explanation:
A For $\mathrm{r} \lt \mathrm{R}$ $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{Gr}$ $\therefore \quad \mathrm{g} \propto \mathrm{r}$ For $r>R$ $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ $\therefore \quad g \propto \frac{1}{r^{2}}$ If $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ If $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}$
UPSEE - 2012
Gravitation
138465
The gravitational potential difference between the surface of a planet and a point $20 \mathrm{~m}$ above it is $16 \mathrm{~J} \mathrm{~kg}^{-1}$. The work done in moving a $4 \mathrm{~kg}$ body by $8 \mathrm{~m}$ on a slope of $60^{\circ}$ from the horizontal is
1 $22.17 \mathrm{~J}$
2 $2.217 \mathrm{~J}$
3 $221.7 \mathrm{~J}$
4 $0.2217 \mathrm{~J}$
Explanation:
A Given that, $\mathrm{h}=20 \mathrm{~m}$ Gravitational Potential difference $=16 \mathrm{~J} / \mathrm{kg}$ Let acceleration due to gravity in the planet is $g_{p}$ Then gravitational potential difference $=\mathrm{g}_{\mathrm{p}} \mathrm{h}$ $16=\mathrm{g}_{\mathrm{p}} \times 20$ $g_{p}=\frac{16}{20}=\frac{4}{5} \mathrm{~m} / \mathrm{s}^{2}$ $\therefore$ Work done $=\mathrm{mg}_{\mathrm{p}} \mathrm{h}$ $=4 \times \frac{4}{5} \times 8 \sin 60^{\circ}$ $=22.17 \mathrm{~J}$
AP EAMCET (22.04.2019) Shift-II
Gravitation
138466
A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is
1 $\frac{m g l}{2}$
2 $\frac{m g R^{2}}{l} \sin \left(\frac{l}{R}\right)$
3 $\frac{m g R^{2}}{l} \sin \left(\frac{R}{l}\right)$
4 $\frac{m g l^{2}}{R} \sin \left(\frac{l}{R}\right)$
Explanation:
B Let $l$ is length of chain $R$ is radius of hemi sphere $\phi=\frac{l}{\mathrm{R}}$ $\mathrm{dm}$ is mass of element $\mathrm{dm}=\frac{\mathrm{mdx}}{l}$ According to figure $\mathrm{h}=\mathrm{R} \cos \theta$ $\mathrm{dE}=\mathrm{dm} \times \mathrm{g} \times \mathrm{h}$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{dx} \times \mathrm{g} \times \mathrm{R} \cos \theta$ According to figure $\mathrm{dx}=\mathrm{Rd} \theta$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{Rd} \theta \times \mathrm{g} \times \mathrm{R} \cos \theta$ $\int_{0}^{\mathrm{E}} \mathrm{dE}=\frac{\mathrm{mgR}^{2}}{l} \int_{0}^{\phi=\frac{l}{\mathrm{R}}} \cos \theta \cdot \mathrm{d} \theta$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l}[\sin \theta]_{0}^{l / \mathrm{R}}$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l} \sin \left(\frac{l}{\mathrm{R}}\right)$
AP EAMCET (23.04.2018) Shift-2
Gravitation
138467
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are arranged in two triangular configurations as shown in figure 1 and figure 2. Work done by an external agent in changing, the configuration from figure 1 to figure 2 is
1 $\frac{6 G m^{2}}{a}\left[2-\frac{6}{\sqrt{2}}\right]$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Gravitation
138464
The magnitudes of gravitational field at distances $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ and $F_{1}$ and $F_{2}$ respectively. Then,
1 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ if $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$
2 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}$ if $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$
3 Both (a) and (b)
4 None of the above
Explanation:
A For $\mathrm{r} \lt \mathrm{R}$ $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{Gr}$ $\therefore \quad \mathrm{g} \propto \mathrm{r}$ For $r>R$ $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ $\therefore \quad g \propto \frac{1}{r^{2}}$ If $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ If $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}$
UPSEE - 2012
Gravitation
138465
The gravitational potential difference between the surface of a planet and a point $20 \mathrm{~m}$ above it is $16 \mathrm{~J} \mathrm{~kg}^{-1}$. The work done in moving a $4 \mathrm{~kg}$ body by $8 \mathrm{~m}$ on a slope of $60^{\circ}$ from the horizontal is
1 $22.17 \mathrm{~J}$
2 $2.217 \mathrm{~J}$
3 $221.7 \mathrm{~J}$
4 $0.2217 \mathrm{~J}$
Explanation:
A Given that, $\mathrm{h}=20 \mathrm{~m}$ Gravitational Potential difference $=16 \mathrm{~J} / \mathrm{kg}$ Let acceleration due to gravity in the planet is $g_{p}$ Then gravitational potential difference $=\mathrm{g}_{\mathrm{p}} \mathrm{h}$ $16=\mathrm{g}_{\mathrm{p}} \times 20$ $g_{p}=\frac{16}{20}=\frac{4}{5} \mathrm{~m} / \mathrm{s}^{2}$ $\therefore$ Work done $=\mathrm{mg}_{\mathrm{p}} \mathrm{h}$ $=4 \times \frac{4}{5} \times 8 \sin 60^{\circ}$ $=22.17 \mathrm{~J}$
AP EAMCET (22.04.2019) Shift-II
Gravitation
138466
A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is
1 $\frac{m g l}{2}$
2 $\frac{m g R^{2}}{l} \sin \left(\frac{l}{R}\right)$
3 $\frac{m g R^{2}}{l} \sin \left(\frac{R}{l}\right)$
4 $\frac{m g l^{2}}{R} \sin \left(\frac{l}{R}\right)$
Explanation:
B Let $l$ is length of chain $R$ is radius of hemi sphere $\phi=\frac{l}{\mathrm{R}}$ $\mathrm{dm}$ is mass of element $\mathrm{dm}=\frac{\mathrm{mdx}}{l}$ According to figure $\mathrm{h}=\mathrm{R} \cos \theta$ $\mathrm{dE}=\mathrm{dm} \times \mathrm{g} \times \mathrm{h}$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{dx} \times \mathrm{g} \times \mathrm{R} \cos \theta$ According to figure $\mathrm{dx}=\mathrm{Rd} \theta$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{Rd} \theta \times \mathrm{g} \times \mathrm{R} \cos \theta$ $\int_{0}^{\mathrm{E}} \mathrm{dE}=\frac{\mathrm{mgR}^{2}}{l} \int_{0}^{\phi=\frac{l}{\mathrm{R}}} \cos \theta \cdot \mathrm{d} \theta$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l}[\sin \theta]_{0}^{l / \mathrm{R}}$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l} \sin \left(\frac{l}{\mathrm{R}}\right)$
AP EAMCET (23.04.2018) Shift-2
Gravitation
138467
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are arranged in two triangular configurations as shown in figure 1 and figure 2. Work done by an external agent in changing, the configuration from figure 1 to figure 2 is
1 $\frac{6 G m^{2}}{a}\left[2-\frac{6}{\sqrt{2}}\right]$
138464
The magnitudes of gravitational field at distances $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ and $F_{1}$ and $F_{2}$ respectively. Then,
1 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ if $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$
2 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}$ if $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$
3 Both (a) and (b)
4 None of the above
Explanation:
A For $\mathrm{r} \lt \mathrm{R}$ $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{Gr}$ $\therefore \quad \mathrm{g} \propto \mathrm{r}$ For $r>R$ $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ $\therefore \quad g \propto \frac{1}{r^{2}}$ If $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ If $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}$
UPSEE - 2012
Gravitation
138465
The gravitational potential difference between the surface of a planet and a point $20 \mathrm{~m}$ above it is $16 \mathrm{~J} \mathrm{~kg}^{-1}$. The work done in moving a $4 \mathrm{~kg}$ body by $8 \mathrm{~m}$ on a slope of $60^{\circ}$ from the horizontal is
1 $22.17 \mathrm{~J}$
2 $2.217 \mathrm{~J}$
3 $221.7 \mathrm{~J}$
4 $0.2217 \mathrm{~J}$
Explanation:
A Given that, $\mathrm{h}=20 \mathrm{~m}$ Gravitational Potential difference $=16 \mathrm{~J} / \mathrm{kg}$ Let acceleration due to gravity in the planet is $g_{p}$ Then gravitational potential difference $=\mathrm{g}_{\mathrm{p}} \mathrm{h}$ $16=\mathrm{g}_{\mathrm{p}} \times 20$ $g_{p}=\frac{16}{20}=\frac{4}{5} \mathrm{~m} / \mathrm{s}^{2}$ $\therefore$ Work done $=\mathrm{mg}_{\mathrm{p}} \mathrm{h}$ $=4 \times \frac{4}{5} \times 8 \sin 60^{\circ}$ $=22.17 \mathrm{~J}$
AP EAMCET (22.04.2019) Shift-II
Gravitation
138466
A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is
1 $\frac{m g l}{2}$
2 $\frac{m g R^{2}}{l} \sin \left(\frac{l}{R}\right)$
3 $\frac{m g R^{2}}{l} \sin \left(\frac{R}{l}\right)$
4 $\frac{m g l^{2}}{R} \sin \left(\frac{l}{R}\right)$
Explanation:
B Let $l$ is length of chain $R$ is radius of hemi sphere $\phi=\frac{l}{\mathrm{R}}$ $\mathrm{dm}$ is mass of element $\mathrm{dm}=\frac{\mathrm{mdx}}{l}$ According to figure $\mathrm{h}=\mathrm{R} \cos \theta$ $\mathrm{dE}=\mathrm{dm} \times \mathrm{g} \times \mathrm{h}$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{dx} \times \mathrm{g} \times \mathrm{R} \cos \theta$ According to figure $\mathrm{dx}=\mathrm{Rd} \theta$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{Rd} \theta \times \mathrm{g} \times \mathrm{R} \cos \theta$ $\int_{0}^{\mathrm{E}} \mathrm{dE}=\frac{\mathrm{mgR}^{2}}{l} \int_{0}^{\phi=\frac{l}{\mathrm{R}}} \cos \theta \cdot \mathrm{d} \theta$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l}[\sin \theta]_{0}^{l / \mathrm{R}}$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l} \sin \left(\frac{l}{\mathrm{R}}\right)$
AP EAMCET (23.04.2018) Shift-2
Gravitation
138467
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are arranged in two triangular configurations as shown in figure 1 and figure 2. Work done by an external agent in changing, the configuration from figure 1 to figure 2 is
1 $\frac{6 G m^{2}}{a}\left[2-\frac{6}{\sqrt{2}}\right]$
138464
The magnitudes of gravitational field at distances $r_{1}$ and $r_{2}$ from the centre of a uniform sphere of radius $R$ and mass $M$ and $F_{1}$ and $F_{2}$ respectively. Then,
1 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ if $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$
2 $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}$ if $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$
3 Both (a) and (b)
4 None of the above
Explanation:
A For $\mathrm{r} \lt \mathrm{R}$ $\mathrm{g}=\frac{4}{3} \pi \rho \mathrm{Gr}$ $\therefore \quad \mathrm{g} \propto \mathrm{r}$ For $r>R$ $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{r}^{2}}$ $\therefore \quad g \propto \frac{1}{r^{2}}$ If $\mathrm{r}_{1} \lt \mathrm{R}$ and $\mathrm{r}_{2} \lt \mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}$ If $\mathrm{r}_{1}>\mathrm{R}$ and $\mathrm{r}_{2}>\mathrm{R}$ then $\frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=\frac{\mathrm{g}_{1}}{\mathrm{~g}_{2}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}$
UPSEE - 2012
Gravitation
138465
The gravitational potential difference between the surface of a planet and a point $20 \mathrm{~m}$ above it is $16 \mathrm{~J} \mathrm{~kg}^{-1}$. The work done in moving a $4 \mathrm{~kg}$ body by $8 \mathrm{~m}$ on a slope of $60^{\circ}$ from the horizontal is
1 $22.17 \mathrm{~J}$
2 $2.217 \mathrm{~J}$
3 $221.7 \mathrm{~J}$
4 $0.2217 \mathrm{~J}$
Explanation:
A Given that, $\mathrm{h}=20 \mathrm{~m}$ Gravitational Potential difference $=16 \mathrm{~J} / \mathrm{kg}$ Let acceleration due to gravity in the planet is $g_{p}$ Then gravitational potential difference $=\mathrm{g}_{\mathrm{p}} \mathrm{h}$ $16=\mathrm{g}_{\mathrm{p}} \times 20$ $g_{p}=\frac{16}{20}=\frac{4}{5} \mathrm{~m} / \mathrm{s}^{2}$ $\therefore$ Work done $=\mathrm{mg}_{\mathrm{p}} \mathrm{h}$ $=4 \times \frac{4}{5} \times 8 \sin 60^{\circ}$ $=22.17 \mathrm{~J}$
AP EAMCET (22.04.2019) Shift-II
Gravitation
138466
A uniform chain of length $l$ and mass $m$ lies on the surface of a smooth hemisphere of radius $R(R>l)$ with one end tied to the top of the hemisphere as shown in the figure. Gravitational potential energy of the chain with respect to the base of the hemisphere is
1 $\frac{m g l}{2}$
2 $\frac{m g R^{2}}{l} \sin \left(\frac{l}{R}\right)$
3 $\frac{m g R^{2}}{l} \sin \left(\frac{R}{l}\right)$
4 $\frac{m g l^{2}}{R} \sin \left(\frac{l}{R}\right)$
Explanation:
B Let $l$ is length of chain $R$ is radius of hemi sphere $\phi=\frac{l}{\mathrm{R}}$ $\mathrm{dm}$ is mass of element $\mathrm{dm}=\frac{\mathrm{mdx}}{l}$ According to figure $\mathrm{h}=\mathrm{R} \cos \theta$ $\mathrm{dE}=\mathrm{dm} \times \mathrm{g} \times \mathrm{h}$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{dx} \times \mathrm{g} \times \mathrm{R} \cos \theta$ According to figure $\mathrm{dx}=\mathrm{Rd} \theta$ $\mathrm{dE}=\frac{\mathrm{m}}{l} \mathrm{Rd} \theta \times \mathrm{g} \times \mathrm{R} \cos \theta$ $\int_{0}^{\mathrm{E}} \mathrm{dE}=\frac{\mathrm{mgR}^{2}}{l} \int_{0}^{\phi=\frac{l}{\mathrm{R}}} \cos \theta \cdot \mathrm{d} \theta$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l}[\sin \theta]_{0}^{l / \mathrm{R}}$ $\mathrm{E}=\frac{\mathrm{mgR}^{2}}{l} \sin \left(\frac{l}{\mathrm{R}}\right)$
AP EAMCET (23.04.2018) Shift-2
Gravitation
138467
Three masses $\mathrm{m}, 2 \mathrm{~m}$ and $3 \mathrm{~m}$ are arranged in two triangular configurations as shown in figure 1 and figure 2. Work done by an external agent in changing, the configuration from figure 1 to figure 2 is
1 $\frac{6 G m^{2}}{a}\left[2-\frac{6}{\sqrt{2}}\right]$
