Is radius of Earth decreased by 10% is the mass remaining unchanged What will happen to the acceleration due to gravity?

Correct Answer: D

Solution :
[d] \[g=\frac{GM}{{{R}^{2}}}\] (i) \[g'=\frac{Gm}{{{\left( \frac{90}{100}R \right)}^{2}}}=\frac{100}{81}-\frac{GM}{{{R}^{2}}}\] (ii) From Eqs. (i) and (ii), \[g'=\frac{100}{81}g\Rightarrow \frac{g'}{g}=\frac{100}{81}\] \[\frac{g'}{g}-1=\frac{100}{81}-1\] \[\therefore \Delta g=\frac{19}{81}g=23%\]of g. So increase is more then 19% of g.

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Correct Answer: C

Solution :
[a] \[\Delta l=\frac{FL}{AY}\] \[\frac{{{(\Delta l)}_{A}}}{{{(\Delta l)}_{B}}}=\left( \frac{{{F}_{A}}}{{{F}_{B}}} \right)\left( \frac{{{L}_{A}}}{{{L}_{B}}} \right)\left( \frac{{{A}_{B}}}{{{A}_{A}}} \right)\left( \frac{{{Y}_{B}}}{{{Y}_{A}}} \right)\] \[=\left( \frac{KM}{3M} \right)(r){{\left( \frac{1}{2r} \right)}^{2}}\left( \frac{1}{3r} \right)=\frac{K}{36{{r}^{2}}}=\frac{1}{6{{r}^{2}}}\](given) K=6 Mass of block P=6M

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Correct Answer: A

Solution :
[a] \[Y=\frac{F/a}{\Delta l/l}=\frac{Fl}{a\Delta l}\] Or \[Y=\frac{Fl\times 4}{\pi {{D}^{2}}\times \Delta l}\] or \[\Delta l\propto \frac{1}{{{D}^{2}}}\]or \[\frac{\Delta {{l}_{2}}}{\Delta {{l}_{1}}}=\frac{{{D}_{1}}^{2}}{{{D}_{2}}^{2}}=\frac{{{n}^{1}}}{1}\]

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Correct Answer: A

Solution :
[a] \[l\propto \frac{1}{Y}\Rightarrow \frac{{{Y}_{s}}}{{{Y}_{c}}}=\frac{{{l}_{c}}}{{{l}_{s}}}\Rightarrow \frac{{{l}_{c}}}{{{l}_{s}}}=\frac{2\times {{10}^{11}}}{1.2\times {{10}^{11}}}=\frac{5}{3}\](i) Also \[{{l}_{c}}-{{l}_{s}}=0.5\] (ii) On solving (i) and (ii), we get \[{{l}_{c}}=0.5\]cm and \[{{l}_{s}}=0.75\]cm

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Correct Answer: D

Solution :
[d] \[\eta =\frac{F/A}{x/L}\Rightarrow x=\frac{L}{\eta }\times \frac{F}{A}\] In \[\eta \]and F are constant, then \[x\propto \frac{L}{A}\] For maximum displacement, area at which force applied should be minimum and vertical side should be maximum. This is given in the R position of rectangular block.

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Correct Answer: B

Solution :
[b] Let L be the original length of the wire and K be the force constant of wire. Final length = initial length + Elongation \[L'=L+\frac{F}{K}\] For first condition \[a=L+\frac{4}{K}\] (i) For second condition \[b=L+\frac{5}{K}\] (ii) By solving equations (i) and (ii), we get \[L=5a-4b\] and \[K=\frac{1}{b-a}\] Now when the longitudinal tension is 9 N, length of the string \[=L+\frac{9}{K}=5a-4b+9(b-a)-5b-4a.\]

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Correct Answer: C

Solution :
[c] \[dF=G\frac{Mdm}{4{{r}^{2}}}\] \[F=\Sigma dF\cos \theta \] \[=\Sigma \frac{GMdm}{4{{r}^{2}}}\cos \theta \] \[=\frac{GM}{4{{r}^{2}}}\times \frac{\sqrt{3r}}{2r}\Sigma dm\] \[=\frac{\sqrt{3}GMm}{8{{r}^{2}}}\] Alternative solution: The gravitational field due to the ring at a distance \[E=\frac{Gm(\sqrt{3r})}{{{[{{r}^{2}}+{{(\sqrt{3}r)}^{2}}]}^{3/2}}}\]or E=\[\frac{\sqrt{3}Gm}{8{{r}^{2}}}\] The required force is EM, i.e., \[(\sqrt{3}Gm)M/8{{r}^{2}}\].

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Correct Answer: D

Solution :
[d] we will be thrown into space, if weight mg is equal to gravitational force duet to the planet. If y is the closest distance. \[\frac{GMm}{{{R}^{2}}}=mg=\frac{G(KM)m}{{{(K'R+y)}^{2}}}\] \[{{(K'R+y)}^{2}}=\frac{KGM}{g}\] \[y={{\left( \frac{KGM}{g} \right)}^{1/2}}-K'R\]

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Correct Answer: B

Solution :
[b] \[{{T}_{O}}=\frac{2\pi r}{\omega }\] \[{{T}_{I}}=\frac{2\pi r}{2\omega }=\frac{{{T}_{O}}}{2}\] Consider an imaginary comet moving along an ellipse. The extreme points of this ellipse are located on orbit of inner planet and the star. Semimajor axis of orbit of such comet will be half of the semi-major axis of the inner planet's orbit. According to Kepler's law, if T is the time period of the comet. \[\frac{T{{'}^{2}}}{{{(r/4)}^{3}}}=\frac{{{T}^{2}}_{I}}{{{(r/2)}^{3}}}\] \[T{{'}^{2}}=\frac{8}{64}{{T}^{2}}_{I}=\frac{{{T}^{2}}_{0}}{32}\] \[(\because {{T}_{1}}={{T}_{0}}/2)\] \[T'=\frac{T}{4\sqrt{2}}\] (T'/2) represents time in which inner planet will fall into star. \[\left( \frac{T'}{2} \right)=\frac{T}{8\sqrt{2}}=\frac{T\sqrt{2}}{16}\]

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Correct Answer: C

Solution :
[c] \[{{T}_{west}}=\frac{2\pi R}{{{v}_{0}}+R\omega }\] and \[{{T}_{east}}=\frac{2\pi R}{{{v}_{0}}-R\omega }\Rightarrow \Delta T={{T}_{east}}\] \[\Rightarrow {{T}_{east}}-{{T}_{west}}=2\pi R\left[ \frac{2\pi R}{{{v}^{2}}_{0}-{{R}^{2}}{{\omega }^{2}}} \right]=\frac{4\pi \omega {{R}^{2}}}{{{v}_{0}}^{2}-{{R}^{2}}{{\omega }^{2}}}\]

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Correct Answer: B

Solution :
[b] Net force on any one particle \[=\frac{G{{M}^{2}}}{{{(2R)}^{2}}}+\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ +\frac{G{{M}^{2}}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ \] \[=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1}{4}+\frac{1}{\sqrt{{}}} \right]\] This force will be equal to centripetal force so \[\frac{M{{u}^{2}}}{R}=\frac{G{{M}^{2}}}{{{R}^{2}}}\left[ \frac{1+2\sqrt{2}}{4} \right]\] \[u=\sqrt{\frac{GM}{R}[1+2\sqrt{2}]}\] \[=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}\]

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Correct Answer: B

Solution :
[b] As gravitational force is a conservative force, work done is independent of path. \[\therefore {{W}_{1}}={{W}_{2}}={{W}_{3}}\]

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Correct Answer: C

Solution :
[c]
\[\frac{G{{M}_{1}}}{{{x}^{2}}}=\frac{G{{M}_{2}}}{{{(R-x)}^{2}}}\] or \[\frac{{{M}_{2}}}{{{M}_{1}}}{{x}^{2}}={{R}^{2}}+{{x}^{2}}-2Rx\] Let \[\frac{{{M}_{2}}}{M}=k\] \[{{x}^{2}}(k-1)+2Rx-{{R}^{2}}=0\] \[x=-\frac{2R+\sqrt{4{{R}^{2}}+4(k-1){{R}^{2}}}}{2(k-1)}=\frac{R\sqrt{{{M}_{1}}}}{\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}}}\] \[R-x=\frac{R\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}}}\] Gravitational potential at point P is \[-\left( \frac{G{{M}_{1}}}{x}+\frac{G{{M}_{2}}}{R-x} \right)\] \[=-\left[ \frac{G{{M}_{1}}(\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}})}{R\sqrt{{{M}_{1}}}}+\frac{G{{M}_{2}}(\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}})}{R\sqrt{{{M}_{2}}}} \right]\] \[=-\left[ \frac{G(\sqrt{{{M}_{2}}}+\sqrt{{{M}_{1}}})}{R}(\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}}) \right]\] \[=-\frac{G{{(\sqrt{{{M}_{1}}}+\sqrt{{{M}_{2}}})}^{2}}}{R}\]