359862
Two particles of masses \(m\) and \(9 m\) are separated by a distance \(r\). At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is \((G=\) universal constant of gravitation)
1 \(-\dfrac{4 G m}{r}\)
2 \(-\dfrac{8 G m}{r}\)
3 \(-\dfrac{16 G m}{r}\)
4 \(-\dfrac{32 G m}{r}\)
Explanation:
Let the gravitational field be zero at a point \('P'\) located on the line joining the two masses, at a distance \(x\) from the smaller mass. The magnitudes of gravitational field intensities by the two masses are equal at \('P'\) \(\begin{aligned}& \Rightarrow \dfrac{G m}{x^{2}}=\dfrac{G(9 m)}{(r-x)^{2}} \\& \Rightarrow 9 x^{2}-(r-x)^{2}=0 \Rightarrow(2 x+r)(4 x-r)=0\end{aligned}\) Neglecting the negative root, \(\Rightarrow x=\dfrac{r}{4}\) Gravitational potential at this point is given by \(\begin{aligned}& \mathrm{V}_{\mathrm{p}}=-\left\{\dfrac{G m}{\dfrac{r}{4}}+\dfrac{G(9 m)}{\dfrac{3 r}{4}}\right\}\left[\because r-x=\dfrac{3 r}{4}\right] \\& =-\left(\dfrac{4 G m+12 G m}{r}\right)=-\left(\dfrac{16 G m}{r}\right)\end{aligned}\)
MHTCET - 2016
PHXI08:GRAVITATION
359863
Two bodies of mass \({10^2}\;kg\) and \({10^3}\;kg\) are lying 1\(m\) apart. The gravitational potential at the midpoint of the line joining them is -
359864
Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a point \(A\) is \( - 5\) unit. If the gravitational potential at point infinite distance away is taken as +10 units, the potential at point \(A\) is
1 +5 unit
2 \({\text{ - 5}}\) unit
3 +15 unit
4 +10 unit
Explanation:
When reference point potential is changed then potential difference between any two points remains constant. \(\begin{aligned}& \Delta V_{\text {Old }}=\Delta V_{\text {New }} \\& \left(V_{\infty}-V_{P}\right)_{\text {Old }}=\left(V_{\infty}-V_{P}\right)_{\text {New }} \\& 0-(-5)=\left(10-V_{P_{\text {New }}}\right) \Rightarrow V_{P_{\text {New }}}=5 \text { unit }\end{aligned}\)
PHXI08:GRAVITATION
359865
Two bodies of masses \(m\) and \(M\) are placed a distance \(d\) apart. The gravitational potential at the position where the gravitational field due to them is zero is
1 \(V=\dfrac{-G}{d}(m+M)\)
2 \(V=\dfrac{-G m}{d}\)
3 \(V=\dfrac{-G M}{d}\)
4 \(V=\dfrac{-G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
Explanation:
If net gravitation field of \(p\) becomes zero means \(\dfrac{G M}{x^{2}}=\dfrac{G M}{(d-x)^{2}} \Rightarrow x=\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}\) and \(d-x=\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}\) Gravitational potential at \(\dfrac{-G M}{\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}}+\dfrac{-G M}{\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}}=-\dfrac{G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
PHXI08:GRAVITATION
359866
Four particles each of mass \(M\), are located at the vertices of a square with side \(L\). The gravitational potential due to this at the centre of the square is
1 \(-\sqrt{64} \dfrac{G M}{L^{2}}\)
2 \(-\sqrt{32} \dfrac{G M}{L}\)
3 \(\sqrt{32} \dfrac{G M}{L}\)
4 Zero
Explanation:
Gravitational potential at the centre is \(U=-4\left(\dfrac{G M}{L / \sqrt{2}}\right)=-\sqrt{32} \dfrac{G M}{L}\)
359862
Two particles of masses \(m\) and \(9 m\) are separated by a distance \(r\). At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is \((G=\) universal constant of gravitation)
1 \(-\dfrac{4 G m}{r}\)
2 \(-\dfrac{8 G m}{r}\)
3 \(-\dfrac{16 G m}{r}\)
4 \(-\dfrac{32 G m}{r}\)
Explanation:
Let the gravitational field be zero at a point \('P'\) located on the line joining the two masses, at a distance \(x\) from the smaller mass. The magnitudes of gravitational field intensities by the two masses are equal at \('P'\) \(\begin{aligned}& \Rightarrow \dfrac{G m}{x^{2}}=\dfrac{G(9 m)}{(r-x)^{2}} \\& \Rightarrow 9 x^{2}-(r-x)^{2}=0 \Rightarrow(2 x+r)(4 x-r)=0\end{aligned}\) Neglecting the negative root, \(\Rightarrow x=\dfrac{r}{4}\) Gravitational potential at this point is given by \(\begin{aligned}& \mathrm{V}_{\mathrm{p}}=-\left\{\dfrac{G m}{\dfrac{r}{4}}+\dfrac{G(9 m)}{\dfrac{3 r}{4}}\right\}\left[\because r-x=\dfrac{3 r}{4}\right] \\& =-\left(\dfrac{4 G m+12 G m}{r}\right)=-\left(\dfrac{16 G m}{r}\right)\end{aligned}\)
MHTCET - 2016
PHXI08:GRAVITATION
359863
Two bodies of mass \({10^2}\;kg\) and \({10^3}\;kg\) are lying 1\(m\) apart. The gravitational potential at the midpoint of the line joining them is -
359864
Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a point \(A\) is \( - 5\) unit. If the gravitational potential at point infinite distance away is taken as +10 units, the potential at point \(A\) is
1 +5 unit
2 \({\text{ - 5}}\) unit
3 +15 unit
4 +10 unit
Explanation:
When reference point potential is changed then potential difference between any two points remains constant. \(\begin{aligned}& \Delta V_{\text {Old }}=\Delta V_{\text {New }} \\& \left(V_{\infty}-V_{P}\right)_{\text {Old }}=\left(V_{\infty}-V_{P}\right)_{\text {New }} \\& 0-(-5)=\left(10-V_{P_{\text {New }}}\right) \Rightarrow V_{P_{\text {New }}}=5 \text { unit }\end{aligned}\)
PHXI08:GRAVITATION
359865
Two bodies of masses \(m\) and \(M\) are placed a distance \(d\) apart. The gravitational potential at the position where the gravitational field due to them is zero is
1 \(V=\dfrac{-G}{d}(m+M)\)
2 \(V=\dfrac{-G m}{d}\)
3 \(V=\dfrac{-G M}{d}\)
4 \(V=\dfrac{-G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
Explanation:
If net gravitation field of \(p\) becomes zero means \(\dfrac{G M}{x^{2}}=\dfrac{G M}{(d-x)^{2}} \Rightarrow x=\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}\) and \(d-x=\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}\) Gravitational potential at \(\dfrac{-G M}{\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}}+\dfrac{-G M}{\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}}=-\dfrac{G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
PHXI08:GRAVITATION
359866
Four particles each of mass \(M\), are located at the vertices of a square with side \(L\). The gravitational potential due to this at the centre of the square is
1 \(-\sqrt{64} \dfrac{G M}{L^{2}}\)
2 \(-\sqrt{32} \dfrac{G M}{L}\)
3 \(\sqrt{32} \dfrac{G M}{L}\)
4 Zero
Explanation:
Gravitational potential at the centre is \(U=-4\left(\dfrac{G M}{L / \sqrt{2}}\right)=-\sqrt{32} \dfrac{G M}{L}\)
359862
Two particles of masses \(m\) and \(9 m\) are separated by a distance \(r\). At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is \((G=\) universal constant of gravitation)
1 \(-\dfrac{4 G m}{r}\)
2 \(-\dfrac{8 G m}{r}\)
3 \(-\dfrac{16 G m}{r}\)
4 \(-\dfrac{32 G m}{r}\)
Explanation:
Let the gravitational field be zero at a point \('P'\) located on the line joining the two masses, at a distance \(x\) from the smaller mass. The magnitudes of gravitational field intensities by the two masses are equal at \('P'\) \(\begin{aligned}& \Rightarrow \dfrac{G m}{x^{2}}=\dfrac{G(9 m)}{(r-x)^{2}} \\& \Rightarrow 9 x^{2}-(r-x)^{2}=0 \Rightarrow(2 x+r)(4 x-r)=0\end{aligned}\) Neglecting the negative root, \(\Rightarrow x=\dfrac{r}{4}\) Gravitational potential at this point is given by \(\begin{aligned}& \mathrm{V}_{\mathrm{p}}=-\left\{\dfrac{G m}{\dfrac{r}{4}}+\dfrac{G(9 m)}{\dfrac{3 r}{4}}\right\}\left[\because r-x=\dfrac{3 r}{4}\right] \\& =-\left(\dfrac{4 G m+12 G m}{r}\right)=-\left(\dfrac{16 G m}{r}\right)\end{aligned}\)
MHTCET - 2016
PHXI08:GRAVITATION
359863
Two bodies of mass \({10^2}\;kg\) and \({10^3}\;kg\) are lying 1\(m\) apart. The gravitational potential at the midpoint of the line joining them is -
359864
Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a point \(A\) is \( - 5\) unit. If the gravitational potential at point infinite distance away is taken as +10 units, the potential at point \(A\) is
1 +5 unit
2 \({\text{ - 5}}\) unit
3 +15 unit
4 +10 unit
Explanation:
When reference point potential is changed then potential difference between any two points remains constant. \(\begin{aligned}& \Delta V_{\text {Old }}=\Delta V_{\text {New }} \\& \left(V_{\infty}-V_{P}\right)_{\text {Old }}=\left(V_{\infty}-V_{P}\right)_{\text {New }} \\& 0-(-5)=\left(10-V_{P_{\text {New }}}\right) \Rightarrow V_{P_{\text {New }}}=5 \text { unit }\end{aligned}\)
PHXI08:GRAVITATION
359865
Two bodies of masses \(m\) and \(M\) are placed a distance \(d\) apart. The gravitational potential at the position where the gravitational field due to them is zero is
1 \(V=\dfrac{-G}{d}(m+M)\)
2 \(V=\dfrac{-G m}{d}\)
3 \(V=\dfrac{-G M}{d}\)
4 \(V=\dfrac{-G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
Explanation:
If net gravitation field of \(p\) becomes zero means \(\dfrac{G M}{x^{2}}=\dfrac{G M}{(d-x)^{2}} \Rightarrow x=\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}\) and \(d-x=\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}\) Gravitational potential at \(\dfrac{-G M}{\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}}+\dfrac{-G M}{\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}}=-\dfrac{G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
PHXI08:GRAVITATION
359866
Four particles each of mass \(M\), are located at the vertices of a square with side \(L\). The gravitational potential due to this at the centre of the square is
1 \(-\sqrt{64} \dfrac{G M}{L^{2}}\)
2 \(-\sqrt{32} \dfrac{G M}{L}\)
3 \(\sqrt{32} \dfrac{G M}{L}\)
4 Zero
Explanation:
Gravitational potential at the centre is \(U=-4\left(\dfrac{G M}{L / \sqrt{2}}\right)=-\sqrt{32} \dfrac{G M}{L}\)
359862
Two particles of masses \(m\) and \(9 m\) are separated by a distance \(r\). At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is \((G=\) universal constant of gravitation)
1 \(-\dfrac{4 G m}{r}\)
2 \(-\dfrac{8 G m}{r}\)
3 \(-\dfrac{16 G m}{r}\)
4 \(-\dfrac{32 G m}{r}\)
Explanation:
Let the gravitational field be zero at a point \('P'\) located on the line joining the two masses, at a distance \(x\) from the smaller mass. The magnitudes of gravitational field intensities by the two masses are equal at \('P'\) \(\begin{aligned}& \Rightarrow \dfrac{G m}{x^{2}}=\dfrac{G(9 m)}{(r-x)^{2}} \\& \Rightarrow 9 x^{2}-(r-x)^{2}=0 \Rightarrow(2 x+r)(4 x-r)=0\end{aligned}\) Neglecting the negative root, \(\Rightarrow x=\dfrac{r}{4}\) Gravitational potential at this point is given by \(\begin{aligned}& \mathrm{V}_{\mathrm{p}}=-\left\{\dfrac{G m}{\dfrac{r}{4}}+\dfrac{G(9 m)}{\dfrac{3 r}{4}}\right\}\left[\because r-x=\dfrac{3 r}{4}\right] \\& =-\left(\dfrac{4 G m+12 G m}{r}\right)=-\left(\dfrac{16 G m}{r}\right)\end{aligned}\)
MHTCET - 2016
PHXI08:GRAVITATION
359863
Two bodies of mass \({10^2}\;kg\) and \({10^3}\;kg\) are lying 1\(m\) apart. The gravitational potential at the midpoint of the line joining them is -
359864
Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a point \(A\) is \( - 5\) unit. If the gravitational potential at point infinite distance away is taken as +10 units, the potential at point \(A\) is
1 +5 unit
2 \({\text{ - 5}}\) unit
3 +15 unit
4 +10 unit
Explanation:
When reference point potential is changed then potential difference between any two points remains constant. \(\begin{aligned}& \Delta V_{\text {Old }}=\Delta V_{\text {New }} \\& \left(V_{\infty}-V_{P}\right)_{\text {Old }}=\left(V_{\infty}-V_{P}\right)_{\text {New }} \\& 0-(-5)=\left(10-V_{P_{\text {New }}}\right) \Rightarrow V_{P_{\text {New }}}=5 \text { unit }\end{aligned}\)
PHXI08:GRAVITATION
359865
Two bodies of masses \(m\) and \(M\) are placed a distance \(d\) apart. The gravitational potential at the position where the gravitational field due to them is zero is
1 \(V=\dfrac{-G}{d}(m+M)\)
2 \(V=\dfrac{-G m}{d}\)
3 \(V=\dfrac{-G M}{d}\)
4 \(V=\dfrac{-G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
Explanation:
If net gravitation field of \(p\) becomes zero means \(\dfrac{G M}{x^{2}}=\dfrac{G M}{(d-x)^{2}} \Rightarrow x=\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}\) and \(d-x=\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}\) Gravitational potential at \(\dfrac{-G M}{\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}}+\dfrac{-G M}{\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}}=-\dfrac{G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
PHXI08:GRAVITATION
359866
Four particles each of mass \(M\), are located at the vertices of a square with side \(L\). The gravitational potential due to this at the centre of the square is
1 \(-\sqrt{64} \dfrac{G M}{L^{2}}\)
2 \(-\sqrt{32} \dfrac{G M}{L}\)
3 \(\sqrt{32} \dfrac{G M}{L}\)
4 Zero
Explanation:
Gravitational potential at the centre is \(U=-4\left(\dfrac{G M}{L / \sqrt{2}}\right)=-\sqrt{32} \dfrac{G M}{L}\)
359862
Two particles of masses \(m\) and \(9 m\) are separated by a distance \(r\). At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is \((G=\) universal constant of gravitation)
1 \(-\dfrac{4 G m}{r}\)
2 \(-\dfrac{8 G m}{r}\)
3 \(-\dfrac{16 G m}{r}\)
4 \(-\dfrac{32 G m}{r}\)
Explanation:
Let the gravitational field be zero at a point \('P'\) located on the line joining the two masses, at a distance \(x\) from the smaller mass. The magnitudes of gravitational field intensities by the two masses are equal at \('P'\) \(\begin{aligned}& \Rightarrow \dfrac{G m}{x^{2}}=\dfrac{G(9 m)}{(r-x)^{2}} \\& \Rightarrow 9 x^{2}-(r-x)^{2}=0 \Rightarrow(2 x+r)(4 x-r)=0\end{aligned}\) Neglecting the negative root, \(\Rightarrow x=\dfrac{r}{4}\) Gravitational potential at this point is given by \(\begin{aligned}& \mathrm{V}_{\mathrm{p}}=-\left\{\dfrac{G m}{\dfrac{r}{4}}+\dfrac{G(9 m)}{\dfrac{3 r}{4}}\right\}\left[\because r-x=\dfrac{3 r}{4}\right] \\& =-\left(\dfrac{4 G m+12 G m}{r}\right)=-\left(\dfrac{16 G m}{r}\right)\end{aligned}\)
MHTCET - 2016
PHXI08:GRAVITATION
359863
Two bodies of mass \({10^2}\;kg\) and \({10^3}\;kg\) are lying 1\(m\) apart. The gravitational potential at the midpoint of the line joining them is -
359864
Taking the gravitational potential at a point infinite distance away as zero, the gravitational potential at a point \(A\) is \( - 5\) unit. If the gravitational potential at point infinite distance away is taken as +10 units, the potential at point \(A\) is
1 +5 unit
2 \({\text{ - 5}}\) unit
3 +15 unit
4 +10 unit
Explanation:
When reference point potential is changed then potential difference between any two points remains constant. \(\begin{aligned}& \Delta V_{\text {Old }}=\Delta V_{\text {New }} \\& \left(V_{\infty}-V_{P}\right)_{\text {Old }}=\left(V_{\infty}-V_{P}\right)_{\text {New }} \\& 0-(-5)=\left(10-V_{P_{\text {New }}}\right) \Rightarrow V_{P_{\text {New }}}=5 \text { unit }\end{aligned}\)
PHXI08:GRAVITATION
359865
Two bodies of masses \(m\) and \(M\) are placed a distance \(d\) apart. The gravitational potential at the position where the gravitational field due to them is zero is
1 \(V=\dfrac{-G}{d}(m+M)\)
2 \(V=\dfrac{-G m}{d}\)
3 \(V=\dfrac{-G M}{d}\)
4 \(V=\dfrac{-G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
Explanation:
If net gravitation field of \(p\) becomes zero means \(\dfrac{G M}{x^{2}}=\dfrac{G M}{(d-x)^{2}} \Rightarrow x=\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}\) and \(d-x=\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}\) Gravitational potential at \(\dfrac{-G M}{\dfrac{(\sqrt{m}) d}{\sqrt{m}+\sqrt{M}}}+\dfrac{-G M}{\dfrac{(\sqrt{M}) d}{\sqrt{m}+\sqrt{M}}}=-\dfrac{G}{d}(\sqrt{m}+\sqrt{M})^{2}\)
PHXI08:GRAVITATION
359866
Four particles each of mass \(M\), are located at the vertices of a square with side \(L\). The gravitational potential due to this at the centre of the square is
1 \(-\sqrt{64} \dfrac{G M}{L^{2}}\)
2 \(-\sqrt{32} \dfrac{G M}{L}\)
3 \(\sqrt{32} \dfrac{G M}{L}\)
4 Zero
Explanation:
Gravitational potential at the centre is \(U=-4\left(\dfrac{G M}{L / \sqrt{2}}\right)=-\sqrt{32} \dfrac{G M}{L}\)
