138483
The gravitational field due to a mass distribution is $E=\frac{k}{x^{3}}$ in the $x$-direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is- [Taking gravitational potential to be zero at infinity]
1 $\frac{\mathrm{k}}{\mathrm{x}}$
2 $\frac{k}{x^{3}}$
3 $\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$
4 $\frac{\mathrm{k}}{3 \mathrm{x}^{3}}$
Explanation:
C According to question, The gravitational field due to a mass distribution in $\mathrm{x}$ direction is- $\mathrm{E}=\frac{\mathrm{k}}{\mathrm{x}^{3}}$ Where, $\mathrm{k}$ is a constant Gravitational force $=\int_{x}^{\infty} E d x$ Putting the value of $E$ from equation (i), we get $=\int_{\mathrm{x}}^{\infty} \frac{\mathrm{k}}{\mathrm{x}^{3}} \mathrm{dx}$ $=\mathrm{k} \int_{\mathrm{x}}^{\infty} \mathrm{x}^{-3} \mathrm{dx}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-3+1}}{-3+1}\right]_{\mathrm{x}}^{\infty}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-2}}{-2}\right]_{\mathrm{x}}^{\infty}$ $=-\frac{\mathrm{k}}{2}\left[\mathrm{x}^{-2}\right]_{\mathrm{x}}^{\infty}$ $=\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$ Hence, the value of gravitational potential at a distance $\mathrm{x}$ is $\mathrm{k} / 2 \mathrm{x}^{2}$.
BCECE-2014
Gravitation
138484
The gravitational potential at a place varies inversely with $x^{2}\left(\right.$ i.e. $\left.V=k / x^{2}\right)$, the gravitational field at that place is
1 $2 \mathrm{k} / \mathrm{x}^{3}$
2 $-2 \mathrm{k} / \mathrm{x}^{3}$
3 $\mathrm{k} / \mathrm{x}$
4 $-\mathrm{k} / \mathrm{x}$
Explanation:
A Given that, Gravitational potential- $\mathrm{V}=\frac{\mathrm{k}}{\mathrm{x}^{2}}$ Gravitational field- $\mathrm{I}=\frac{-\mathrm{d}}{\mathrm{dx}}(\mathrm{V})$ Putting the value of $\mathrm{V}$ from equation (i), we get $I=\frac{-d}{d x}\left(\frac{k}{x^{2}}\right)$ $=\frac{-\mathrm{d}}{\mathrm{dx}}\left(\mathrm{kx}^{-2}\right)$ $=-\mathrm{k} \times \frac{-2}{\mathrm{x}^{3}}$ $\mathrm{I}=\frac{2 \mathrm{k}}{\mathrm{x}^{3}}$
VITEEE-2013
Gravitation
138485
A projectile is fired from the earth vertically with a velocity $\mathrm{kv}$ where $\mathrm{v}$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is : Where $\mathrm{k} \lt 1$.
1 $\frac{\mathrm{R}}{\mathrm{k}^{2}}$
2 $\frac{\mathrm{R}}{1+\mathrm{k}^{2}}$
3 $\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
4 $\frac{\mathrm{kR}}{1+\mathrm{k}^{2}}$
Explanation:
C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $\mathrm{h}$, then according to the law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$ Here, $\mathrm{v}=\mathrm{kv}_{\mathrm{e}}$ $\mathrm{v}=\mathrm{k} \sqrt{2 \mathrm{gR}}$ $\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\right)$ Squaring both side, $\mathrm{v}^{2}=2 \mathrm{k}^{2} \mathrm{gR}$ Putting the value of $v^{2}$ in equation (i), we get $\frac{1}{2} \mathrm{mk}^{2} 2 \mathrm{gR}=\frac{\mathrm{mg}(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]} \quad\{\mathrm{h}=\mathrm{r}-\mathrm{R}\}$ $\mathrm{k}^{2} \mathrm{R}=\frac{(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]}$ $\mathrm{k}^{2} \mathrm{R}\left[1+\frac{\mathrm{r}-\mathrm{R}}{\mathrm{R}}\right]=\mathrm{r}-\mathrm{R}$ or $k^{2} r=r-R$ $\mathrm{R}=\mathrm{r}-\mathrm{k}^{2} \mathrm{r}$ $\mathrm{R}=\mathrm{r}\left(1-\mathrm{k}^{2}\right)$ $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
SCRA-2012
Gravitation
138486
A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :
1 $\frac{\mathrm{mgR}}{2}$
2 $\mathrm{mgR}$
3 $2 \mathrm{mgR}$
4 $\frac{\operatorname{mgR}}{4}$
Explanation:
A At Earth surface gravitational potential energy of the mass $\mathrm{U}_{\mathrm{E}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ At target gravitational force- $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{h}+\mathrm{R}} \quad(\because \mathrm{h}=\mathrm{R})$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}$ From equation (i) and (ii), we get Work done $(\mathrm{W})=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)$ $=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}=\frac{\mathrm{GMm}}{2 \mathrm{R}}\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\therefore \mathrm{W}=\frac{\mathrm{mgR}}{2}$
138483
The gravitational field due to a mass distribution is $E=\frac{k}{x^{3}}$ in the $x$-direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is- [Taking gravitational potential to be zero at infinity]
1 $\frac{\mathrm{k}}{\mathrm{x}}$
2 $\frac{k}{x^{3}}$
3 $\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$
4 $\frac{\mathrm{k}}{3 \mathrm{x}^{3}}$
Explanation:
C According to question, The gravitational field due to a mass distribution in $\mathrm{x}$ direction is- $\mathrm{E}=\frac{\mathrm{k}}{\mathrm{x}^{3}}$ Where, $\mathrm{k}$ is a constant Gravitational force $=\int_{x}^{\infty} E d x$ Putting the value of $E$ from equation (i), we get $=\int_{\mathrm{x}}^{\infty} \frac{\mathrm{k}}{\mathrm{x}^{3}} \mathrm{dx}$ $=\mathrm{k} \int_{\mathrm{x}}^{\infty} \mathrm{x}^{-3} \mathrm{dx}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-3+1}}{-3+1}\right]_{\mathrm{x}}^{\infty}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-2}}{-2}\right]_{\mathrm{x}}^{\infty}$ $=-\frac{\mathrm{k}}{2}\left[\mathrm{x}^{-2}\right]_{\mathrm{x}}^{\infty}$ $=\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$ Hence, the value of gravitational potential at a distance $\mathrm{x}$ is $\mathrm{k} / 2 \mathrm{x}^{2}$.
BCECE-2014
Gravitation
138484
The gravitational potential at a place varies inversely with $x^{2}\left(\right.$ i.e. $\left.V=k / x^{2}\right)$, the gravitational field at that place is
1 $2 \mathrm{k} / \mathrm{x}^{3}$
2 $-2 \mathrm{k} / \mathrm{x}^{3}$
3 $\mathrm{k} / \mathrm{x}$
4 $-\mathrm{k} / \mathrm{x}$
Explanation:
A Given that, Gravitational potential- $\mathrm{V}=\frac{\mathrm{k}}{\mathrm{x}^{2}}$ Gravitational field- $\mathrm{I}=\frac{-\mathrm{d}}{\mathrm{dx}}(\mathrm{V})$ Putting the value of $\mathrm{V}$ from equation (i), we get $I=\frac{-d}{d x}\left(\frac{k}{x^{2}}\right)$ $=\frac{-\mathrm{d}}{\mathrm{dx}}\left(\mathrm{kx}^{-2}\right)$ $=-\mathrm{k} \times \frac{-2}{\mathrm{x}^{3}}$ $\mathrm{I}=\frac{2 \mathrm{k}}{\mathrm{x}^{3}}$
VITEEE-2013
Gravitation
138485
A projectile is fired from the earth vertically with a velocity $\mathrm{kv}$ where $\mathrm{v}$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is : Where $\mathrm{k} \lt 1$.
1 $\frac{\mathrm{R}}{\mathrm{k}^{2}}$
2 $\frac{\mathrm{R}}{1+\mathrm{k}^{2}}$
3 $\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
4 $\frac{\mathrm{kR}}{1+\mathrm{k}^{2}}$
Explanation:
C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $\mathrm{h}$, then according to the law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$ Here, $\mathrm{v}=\mathrm{kv}_{\mathrm{e}}$ $\mathrm{v}=\mathrm{k} \sqrt{2 \mathrm{gR}}$ $\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\right)$ Squaring both side, $\mathrm{v}^{2}=2 \mathrm{k}^{2} \mathrm{gR}$ Putting the value of $v^{2}$ in equation (i), we get $\frac{1}{2} \mathrm{mk}^{2} 2 \mathrm{gR}=\frac{\mathrm{mg}(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]} \quad\{\mathrm{h}=\mathrm{r}-\mathrm{R}\}$ $\mathrm{k}^{2} \mathrm{R}=\frac{(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]}$ $\mathrm{k}^{2} \mathrm{R}\left[1+\frac{\mathrm{r}-\mathrm{R}}{\mathrm{R}}\right]=\mathrm{r}-\mathrm{R}$ or $k^{2} r=r-R$ $\mathrm{R}=\mathrm{r}-\mathrm{k}^{2} \mathrm{r}$ $\mathrm{R}=\mathrm{r}\left(1-\mathrm{k}^{2}\right)$ $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
SCRA-2012
Gravitation
138486
A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :
1 $\frac{\mathrm{mgR}}{2}$
2 $\mathrm{mgR}$
3 $2 \mathrm{mgR}$
4 $\frac{\operatorname{mgR}}{4}$
Explanation:
A At Earth surface gravitational potential energy of the mass $\mathrm{U}_{\mathrm{E}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ At target gravitational force- $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{h}+\mathrm{R}} \quad(\because \mathrm{h}=\mathrm{R})$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}$ From equation (i) and (ii), we get Work done $(\mathrm{W})=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)$ $=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}=\frac{\mathrm{GMm}}{2 \mathrm{R}}\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\therefore \mathrm{W}=\frac{\mathrm{mgR}}{2}$
138483
The gravitational field due to a mass distribution is $E=\frac{k}{x^{3}}$ in the $x$-direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is- [Taking gravitational potential to be zero at infinity]
1 $\frac{\mathrm{k}}{\mathrm{x}}$
2 $\frac{k}{x^{3}}$
3 $\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$
4 $\frac{\mathrm{k}}{3 \mathrm{x}^{3}}$
Explanation:
C According to question, The gravitational field due to a mass distribution in $\mathrm{x}$ direction is- $\mathrm{E}=\frac{\mathrm{k}}{\mathrm{x}^{3}}$ Where, $\mathrm{k}$ is a constant Gravitational force $=\int_{x}^{\infty} E d x$ Putting the value of $E$ from equation (i), we get $=\int_{\mathrm{x}}^{\infty} \frac{\mathrm{k}}{\mathrm{x}^{3}} \mathrm{dx}$ $=\mathrm{k} \int_{\mathrm{x}}^{\infty} \mathrm{x}^{-3} \mathrm{dx}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-3+1}}{-3+1}\right]_{\mathrm{x}}^{\infty}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-2}}{-2}\right]_{\mathrm{x}}^{\infty}$ $=-\frac{\mathrm{k}}{2}\left[\mathrm{x}^{-2}\right]_{\mathrm{x}}^{\infty}$ $=\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$ Hence, the value of gravitational potential at a distance $\mathrm{x}$ is $\mathrm{k} / 2 \mathrm{x}^{2}$.
BCECE-2014
Gravitation
138484
The gravitational potential at a place varies inversely with $x^{2}\left(\right.$ i.e. $\left.V=k / x^{2}\right)$, the gravitational field at that place is
1 $2 \mathrm{k} / \mathrm{x}^{3}$
2 $-2 \mathrm{k} / \mathrm{x}^{3}$
3 $\mathrm{k} / \mathrm{x}$
4 $-\mathrm{k} / \mathrm{x}$
Explanation:
A Given that, Gravitational potential- $\mathrm{V}=\frac{\mathrm{k}}{\mathrm{x}^{2}}$ Gravitational field- $\mathrm{I}=\frac{-\mathrm{d}}{\mathrm{dx}}(\mathrm{V})$ Putting the value of $\mathrm{V}$ from equation (i), we get $I=\frac{-d}{d x}\left(\frac{k}{x^{2}}\right)$ $=\frac{-\mathrm{d}}{\mathrm{dx}}\left(\mathrm{kx}^{-2}\right)$ $=-\mathrm{k} \times \frac{-2}{\mathrm{x}^{3}}$ $\mathrm{I}=\frac{2 \mathrm{k}}{\mathrm{x}^{3}}$
VITEEE-2013
Gravitation
138485
A projectile is fired from the earth vertically with a velocity $\mathrm{kv}$ where $\mathrm{v}$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is : Where $\mathrm{k} \lt 1$.
1 $\frac{\mathrm{R}}{\mathrm{k}^{2}}$
2 $\frac{\mathrm{R}}{1+\mathrm{k}^{2}}$
3 $\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
4 $\frac{\mathrm{kR}}{1+\mathrm{k}^{2}}$
Explanation:
C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $\mathrm{h}$, then according to the law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$ Here, $\mathrm{v}=\mathrm{kv}_{\mathrm{e}}$ $\mathrm{v}=\mathrm{k} \sqrt{2 \mathrm{gR}}$ $\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\right)$ Squaring both side, $\mathrm{v}^{2}=2 \mathrm{k}^{2} \mathrm{gR}$ Putting the value of $v^{2}$ in equation (i), we get $\frac{1}{2} \mathrm{mk}^{2} 2 \mathrm{gR}=\frac{\mathrm{mg}(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]} \quad\{\mathrm{h}=\mathrm{r}-\mathrm{R}\}$ $\mathrm{k}^{2} \mathrm{R}=\frac{(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]}$ $\mathrm{k}^{2} \mathrm{R}\left[1+\frac{\mathrm{r}-\mathrm{R}}{\mathrm{R}}\right]=\mathrm{r}-\mathrm{R}$ or $k^{2} r=r-R$ $\mathrm{R}=\mathrm{r}-\mathrm{k}^{2} \mathrm{r}$ $\mathrm{R}=\mathrm{r}\left(1-\mathrm{k}^{2}\right)$ $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
SCRA-2012
Gravitation
138486
A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :
1 $\frac{\mathrm{mgR}}{2}$
2 $\mathrm{mgR}$
3 $2 \mathrm{mgR}$
4 $\frac{\operatorname{mgR}}{4}$
Explanation:
A At Earth surface gravitational potential energy of the mass $\mathrm{U}_{\mathrm{E}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ At target gravitational force- $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{h}+\mathrm{R}} \quad(\because \mathrm{h}=\mathrm{R})$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}$ From equation (i) and (ii), we get Work done $(\mathrm{W})=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)$ $=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}=\frac{\mathrm{GMm}}{2 \mathrm{R}}\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\therefore \mathrm{W}=\frac{\mathrm{mgR}}{2}$
138483
The gravitational field due to a mass distribution is $E=\frac{k}{x^{3}}$ in the $x$-direction, where $k$ is a constant. The value of gravitational potential at a distance $x$ is- [Taking gravitational potential to be zero at infinity]
1 $\frac{\mathrm{k}}{\mathrm{x}}$
2 $\frac{k}{x^{3}}$
3 $\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$
4 $\frac{\mathrm{k}}{3 \mathrm{x}^{3}}$
Explanation:
C According to question, The gravitational field due to a mass distribution in $\mathrm{x}$ direction is- $\mathrm{E}=\frac{\mathrm{k}}{\mathrm{x}^{3}}$ Where, $\mathrm{k}$ is a constant Gravitational force $=\int_{x}^{\infty} E d x$ Putting the value of $E$ from equation (i), we get $=\int_{\mathrm{x}}^{\infty} \frac{\mathrm{k}}{\mathrm{x}^{3}} \mathrm{dx}$ $=\mathrm{k} \int_{\mathrm{x}}^{\infty} \mathrm{x}^{-3} \mathrm{dx}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-3+1}}{-3+1}\right]_{\mathrm{x}}^{\infty}$ $=\mathrm{k}\left[\frac{\mathrm{x}^{-2}}{-2}\right]_{\mathrm{x}}^{\infty}$ $=-\frac{\mathrm{k}}{2}\left[\mathrm{x}^{-2}\right]_{\mathrm{x}}^{\infty}$ $=\frac{\mathrm{k}}{2 \mathrm{x}^{2}}$ Hence, the value of gravitational potential at a distance $\mathrm{x}$ is $\mathrm{k} / 2 \mathrm{x}^{2}$.
BCECE-2014
Gravitation
138484
The gravitational potential at a place varies inversely with $x^{2}\left(\right.$ i.e. $\left.V=k / x^{2}\right)$, the gravitational field at that place is
1 $2 \mathrm{k} / \mathrm{x}^{3}$
2 $-2 \mathrm{k} / \mathrm{x}^{3}$
3 $\mathrm{k} / \mathrm{x}$
4 $-\mathrm{k} / \mathrm{x}$
Explanation:
A Given that, Gravitational potential- $\mathrm{V}=\frac{\mathrm{k}}{\mathrm{x}^{2}}$ Gravitational field- $\mathrm{I}=\frac{-\mathrm{d}}{\mathrm{dx}}(\mathrm{V})$ Putting the value of $\mathrm{V}$ from equation (i), we get $I=\frac{-d}{d x}\left(\frac{k}{x^{2}}\right)$ $=\frac{-\mathrm{d}}{\mathrm{dx}}\left(\mathrm{kx}^{-2}\right)$ $=-\mathrm{k} \times \frac{-2}{\mathrm{x}^{3}}$ $\mathrm{I}=\frac{2 \mathrm{k}}{\mathrm{x}^{3}}$
VITEEE-2013
Gravitation
138485
A projectile is fired from the earth vertically with a velocity $\mathrm{kv}$ where $\mathrm{v}$ is the escape velocity and $k$ is constant. The maximum height to which it rises as measured from the centre of earth of radius $R$ is : Where $\mathrm{k} \lt 1$.
1 $\frac{\mathrm{R}}{\mathrm{k}^{2}}$
2 $\frac{\mathrm{R}}{1+\mathrm{k}^{2}}$
3 $\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
4 $\frac{\mathrm{kR}}{1+\mathrm{k}^{2}}$
Explanation:
C According to question, if a body is projected from the surface of the earth with a velocity $v$ and reaches at height $\mathrm{h}$, then according to the law of conservation of energy, $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}$ Here, $\mathrm{v}=\mathrm{kv}_{\mathrm{e}}$ $\mathrm{v}=\mathrm{k} \sqrt{2 \mathrm{gR}}$ $\left(\because \mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{gR}}\right)$ Squaring both side, $\mathrm{v}^{2}=2 \mathrm{k}^{2} \mathrm{gR}$ Putting the value of $v^{2}$ in equation (i), we get $\frac{1}{2} \mathrm{mk}^{2} 2 \mathrm{gR}=\frac{\mathrm{mg}(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]} \quad\{\mathrm{h}=\mathrm{r}-\mathrm{R}\}$ $\mathrm{k}^{2} \mathrm{R}=\frac{(\mathrm{r}-\mathrm{R})}{\left[1+\frac{(\mathrm{r}-\mathrm{R})}{\mathrm{R}}\right]}$ $\mathrm{k}^{2} \mathrm{R}\left[1+\frac{\mathrm{r}-\mathrm{R}}{\mathrm{R}}\right]=\mathrm{r}-\mathrm{R}$ or $k^{2} r=r-R$ $\mathrm{R}=\mathrm{r}-\mathrm{k}^{2} \mathrm{r}$ $\mathrm{R}=\mathrm{r}\left(1-\mathrm{k}^{2}\right)$ $\mathrm{r}=\frac{\mathrm{R}}{1-\mathrm{k}^{2}}$
SCRA-2012
Gravitation
138486
A mass $m$ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If $R$ is the radius and $M$ is the mass of the Earth, then work done in this process is :
1 $\frac{\mathrm{mgR}}{2}$
2 $\mathrm{mgR}$
3 $2 \mathrm{mgR}$
4 $\frac{\operatorname{mgR}}{4}$
Explanation:
A At Earth surface gravitational potential energy of the mass $\mathrm{U}_{\mathrm{E}}=\frac{-\mathrm{GMm}}{\mathrm{R}}$ At target gravitational force- $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{h}+\mathrm{R}} \quad(\because \mathrm{h}=\mathrm{R})$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{\mathrm{R}+\mathrm{R}}$ $\mathrm{U}_{\mathrm{T}}=\frac{-\mathrm{GMm}}{2 \mathrm{R}}$ From equation (i) and (ii), we get Work done $(\mathrm{W})=\frac{-\mathrm{GMm}}{2 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right)$ $=\frac{-\mathrm{GMm}+2 \mathrm{GMm}}{2 \mathrm{R}}=\frac{\mathrm{GMm}}{2 \mathrm{R}}\left(\because \mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} \Rightarrow \frac{\mathrm{GM}}{\mathrm{R}}=\mathrm{gR}\right)$ $\therefore \mathrm{W}=\frac{\mathrm{mgR}}{2}$
