NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI08:GRAVITATION
359859
If a spherical shell is cut into two pieces along a chord as shown in figure and point \(P\) and \(Q\) have gravitational intensities \(I_{P}\) and \(I_{Q}\) respectively, then
1 \(I_{P}>I_{Q}\)
2 \(I_{P}=I_{Q}=0\)
3 \(I_{P}=I_{Q} \neq 0\)
4 \(I_{P} < I_{Q}\)
Explanation:
We know that inside a spherical shell gravitational field intensity at any point is zero, hence \(\begin{aligned}& I_{P}+I_{Q}=0 \\& \Rightarrow \quad I_{P}=I_{Q} \neq 0\end{aligned}\)
PHXI08:GRAVITATION
359860
A ring has mass \(M\), radius \(R\). A point mass \(m\) is placed at a distance \(x\) on the axial line as shown. Find \(x\) so that force experienced is maximum.
1 \(\dfrac{R}{3}\)
2 \(\dfrac{R}{2}\)
3 \(\dfrac{R}{\sqrt{3}}\)
4 \(\dfrac{R}{\sqrt{2}}\)
Explanation:
The force exerted by the ring on the point mass is \(F=\dfrac{G M m x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\) When the force \(F\) is maximum \(\begin{aligned}& \dfrac{d F}{d x}=G M m \dfrac{d}{d x}\left(\dfrac{x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\right)=0 \\& \Rightarrow \dfrac{x \dfrac{d}{d x}\left(R^{2}+x^{2}\right)^{3 / 2}-\left(R^{2}+x^{2}\right)^{3 / 2} \dfrac{d}{d x}(x)}{\left(R^{2}+x^{2}\right)^{3 / 2}}=0 \\& x \dfrac{3}{2}\left(R^{2}+x^{2}\right)^{1 / 2}(2 x)-\left(R^{2}+x^{2}\right)^{3 / 2}=0 \\& 3 x^{2}-\left(R^{2}+x^{2}\right)=0 \\& \Rightarrow \quad x=\dfrac{+R}{\sqrt{2}}\end{aligned}\)
PHXI08:GRAVITATION
359861
Gravitational field intensity at the centre of the semi circle formed by a thin wire \(A B\) of mass ' \(m\) ' and length ' \(L\) ' is
1 \(\dfrac{G m^{2}}{L^{2}}(\hat{i})\)
2 \(\dfrac{G m^{2}}{\pi L^{2}}(\hat{j})\)
3 \(\dfrac{2 \pi G m}{L^{2}}(\hat{i})\)
4 \(\dfrac{2 \pi G m}{L^{2}}(\hat{j})\)
Explanation:
\(\lambda=\dfrac{m}{L} ; L=\pi r ; d m=\lambda d l=\lambda(r d \theta)\) \(\begin{aligned}& E=\dfrac{G \lambda}{r}\left[\int_{0}^{\pi} \cos \theta d \theta \hat{i}+\int_{0}^{\pi} \sin \theta d \theta \hat{j}\right] \\& \vec{E}=\dfrac{G \lambda}{r}\left[(\sin \theta)_{0}^{\pi} \hat{i}+(-\cos \theta)_{0}^{\pi} \hat{j}\right]\end{aligned}\) \(\begin{aligned}\Rightarrow \vec{E} & =\dfrac{2 G m}{r} \hat{j} \\& =\dfrac{2 G \lambda}{L \dfrac{L}{\pi}} \hat{j}\end{aligned}\) \(=\dfrac{2 \pi G m}{L^{2}} \hat{j}\)
359859
If a spherical shell is cut into two pieces along a chord as shown in figure and point \(P\) and \(Q\) have gravitational intensities \(I_{P}\) and \(I_{Q}\) respectively, then
1 \(I_{P}>I_{Q}\)
2 \(I_{P}=I_{Q}=0\)
3 \(I_{P}=I_{Q} \neq 0\)
4 \(I_{P} < I_{Q}\)
Explanation:
We know that inside a spherical shell gravitational field intensity at any point is zero, hence \(\begin{aligned}& I_{P}+I_{Q}=0 \\& \Rightarrow \quad I_{P}=I_{Q} \neq 0\end{aligned}\)
PHXI08:GRAVITATION
359860
A ring has mass \(M\), radius \(R\). A point mass \(m\) is placed at a distance \(x\) on the axial line as shown. Find \(x\) so that force experienced is maximum.
1 \(\dfrac{R}{3}\)
2 \(\dfrac{R}{2}\)
3 \(\dfrac{R}{\sqrt{3}}\)
4 \(\dfrac{R}{\sqrt{2}}\)
Explanation:
The force exerted by the ring on the point mass is \(F=\dfrac{G M m x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\) When the force \(F\) is maximum \(\begin{aligned}& \dfrac{d F}{d x}=G M m \dfrac{d}{d x}\left(\dfrac{x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\right)=0 \\& \Rightarrow \dfrac{x \dfrac{d}{d x}\left(R^{2}+x^{2}\right)^{3 / 2}-\left(R^{2}+x^{2}\right)^{3 / 2} \dfrac{d}{d x}(x)}{\left(R^{2}+x^{2}\right)^{3 / 2}}=0 \\& x \dfrac{3}{2}\left(R^{2}+x^{2}\right)^{1 / 2}(2 x)-\left(R^{2}+x^{2}\right)^{3 / 2}=0 \\& 3 x^{2}-\left(R^{2}+x^{2}\right)=0 \\& \Rightarrow \quad x=\dfrac{+R}{\sqrt{2}}\end{aligned}\)
PHXI08:GRAVITATION
359861
Gravitational field intensity at the centre of the semi circle formed by a thin wire \(A B\) of mass ' \(m\) ' and length ' \(L\) ' is
1 \(\dfrac{G m^{2}}{L^{2}}(\hat{i})\)
2 \(\dfrac{G m^{2}}{\pi L^{2}}(\hat{j})\)
3 \(\dfrac{2 \pi G m}{L^{2}}(\hat{i})\)
4 \(\dfrac{2 \pi G m}{L^{2}}(\hat{j})\)
Explanation:
\(\lambda=\dfrac{m}{L} ; L=\pi r ; d m=\lambda d l=\lambda(r d \theta)\) \(\begin{aligned}& E=\dfrac{G \lambda}{r}\left[\int_{0}^{\pi} \cos \theta d \theta \hat{i}+\int_{0}^{\pi} \sin \theta d \theta \hat{j}\right] \\& \vec{E}=\dfrac{G \lambda}{r}\left[(\sin \theta)_{0}^{\pi} \hat{i}+(-\cos \theta)_{0}^{\pi} \hat{j}\right]\end{aligned}\) \(\begin{aligned}\Rightarrow \vec{E} & =\dfrac{2 G m}{r} \hat{j} \\& =\dfrac{2 G \lambda}{L \dfrac{L}{\pi}} \hat{j}\end{aligned}\) \(=\dfrac{2 \pi G m}{L^{2}} \hat{j}\)
359859
If a spherical shell is cut into two pieces along a chord as shown in figure and point \(P\) and \(Q\) have gravitational intensities \(I_{P}\) and \(I_{Q}\) respectively, then
1 \(I_{P}>I_{Q}\)
2 \(I_{P}=I_{Q}=0\)
3 \(I_{P}=I_{Q} \neq 0\)
4 \(I_{P} < I_{Q}\)
Explanation:
We know that inside a spherical shell gravitational field intensity at any point is zero, hence \(\begin{aligned}& I_{P}+I_{Q}=0 \\& \Rightarrow \quad I_{P}=I_{Q} \neq 0\end{aligned}\)
PHXI08:GRAVITATION
359860
A ring has mass \(M\), radius \(R\). A point mass \(m\) is placed at a distance \(x\) on the axial line as shown. Find \(x\) so that force experienced is maximum.
1 \(\dfrac{R}{3}\)
2 \(\dfrac{R}{2}\)
3 \(\dfrac{R}{\sqrt{3}}\)
4 \(\dfrac{R}{\sqrt{2}}\)
Explanation:
The force exerted by the ring on the point mass is \(F=\dfrac{G M m x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\) When the force \(F\) is maximum \(\begin{aligned}& \dfrac{d F}{d x}=G M m \dfrac{d}{d x}\left(\dfrac{x}{\left(R^{2}+x^{2}\right)^{3 / 2}}\right)=0 \\& \Rightarrow \dfrac{x \dfrac{d}{d x}\left(R^{2}+x^{2}\right)^{3 / 2}-\left(R^{2}+x^{2}\right)^{3 / 2} \dfrac{d}{d x}(x)}{\left(R^{2}+x^{2}\right)^{3 / 2}}=0 \\& x \dfrac{3}{2}\left(R^{2}+x^{2}\right)^{1 / 2}(2 x)-\left(R^{2}+x^{2}\right)^{3 / 2}=0 \\& 3 x^{2}-\left(R^{2}+x^{2}\right)=0 \\& \Rightarrow \quad x=\dfrac{+R}{\sqrt{2}}\end{aligned}\)
PHXI08:GRAVITATION
359861
Gravitational field intensity at the centre of the semi circle formed by a thin wire \(A B\) of mass ' \(m\) ' and length ' \(L\) ' is
1 \(\dfrac{G m^{2}}{L^{2}}(\hat{i})\)
2 \(\dfrac{G m^{2}}{\pi L^{2}}(\hat{j})\)
3 \(\dfrac{2 \pi G m}{L^{2}}(\hat{i})\)
4 \(\dfrac{2 \pi G m}{L^{2}}(\hat{j})\)
Explanation:
\(\lambda=\dfrac{m}{L} ; L=\pi r ; d m=\lambda d l=\lambda(r d \theta)\) \(\begin{aligned}& E=\dfrac{G \lambda}{r}\left[\int_{0}^{\pi} \cos \theta d \theta \hat{i}+\int_{0}^{\pi} \sin \theta d \theta \hat{j}\right] \\& \vec{E}=\dfrac{G \lambda}{r}\left[(\sin \theta)_{0}^{\pi} \hat{i}+(-\cos \theta)_{0}^{\pi} \hat{j}\right]\end{aligned}\) \(\begin{aligned}\Rightarrow \vec{E} & =\dfrac{2 G m}{r} \hat{j} \\& =\dfrac{2 G \lambda}{L \dfrac{L}{\pi}} \hat{j}\end{aligned}\) \(=\dfrac{2 \pi G m}{L^{2}} \hat{j}\)
