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Correct Answer: B
Solution :
\[{{S}_{n}}th=u+\frac{a}{2}\left( 2n-1 \right)\] \[\text{So, }\frac{{{S}_{4}}}{{{S}_{5}}}=\frac{0+\frac{g}{2}\left( 7 \right)}{0+\frac{g}{2}\left( 9 \right)}=\frac{7}{9}\]
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Correct Answer: A
Solution :
\[s=ut+\frac{1}{2}a{{t}^{2}}\] \[-65=12t-5{{t}^{2}}\]on solving we get, t=5s
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Correct Answer: D
Solution :
For A to B \[S=\frac{1}{2}g{{t}^{2}}\text{ }....\text{(i)}\] For A to C \[\text{S=}\frac{1}{2}gt{{'}^{2}}\text{ }....\text{(ii)}\] Dividing (i) by (ii) we get \[\frac{t}{t'}=\frac{1}{\sqrt{2}}\]
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Page 4
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Correct Answer: A
Solution :
Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s. Now, distance moved in 18 s by 1st ball \[=\frac{1}{2}\times 10\times {{18}^{2}}=90\times 18=1620\text{ m }\] Distance moved in 12 s by 2nd ball \[=ut+\frac{1}{2}g{{t}^{2}}\text{ }\therefore \text{1620=12v+5}\times \text{144}\] \[\Rightarrow \text{v=135-60=75 m}{{\text{s}}^{-1}}\]
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Correct Answer: A
Solution :
\[\therefore h=\frac{1}{2}g{{t}^{2}}\text{ }\therefore {{\text{h}}_{1}}=\frac{1}{2}g{{\left( 5 \right)}^{2}}=125\] \[{{\text{h}}_{1}}+{{\text{h}}_{2}}=\frac{1}{2}g{{\left( 10 \right)}^{2}}=500\Rightarrow {{\text{h}}_{2}}=375\] \[\therefore {{\text{h}}_{1}}+{{\text{h}}_{2}}+{{h}_{3}}=\frac{1}{2}g{{\left( 15 \right)}^{2}}=1125\] \[\Rightarrow {{h}_{3}}625\] \[{{h}_{2}}=3{{h}_{1}},{{h}_{3}}=5{{h}_{1}}\text{ or }{{\text{h}}_{1}}=\frac{{{h}_{2}}}{3}=\frac{{{h}_{3}}}{5}\]
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Correct Answer: C
Solution :
In \[\frac{\text{T}}{\text{3}}\] sec, the distance travelled =\[\frac{\text{1}}{\text{2}}\text{g}{{\left( \frac{\text{T}}{\text{3}} \right)}^{\text{2}}}\text{=}\frac{\text{h}}{\text{9}}\] \[\therefore \]Position of the ball from the ground =\[h-\frac{h}{9}=\frac{8h}{9}\text{m}\]
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Page 8
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Correct Answer: B
Solution :
Time taken by the stone to reach the water level \[{{t}_{1}}=\sqrt{\frac{2h}{g}}\] Time taken by sound to come to the mouth of the well, \[{{\text{t}}_{\text{2}}}\text{=}\frac{\text{h}}{\text{v}}\] \[\therefore \]Total time \[{{\text{t}}_{\text{1}}}\text{+}{{\text{t}}_{\text{2}}}\text{=}\sqrt{\frac{\text{2h}}{\text{g}}}\text{+}\frac{\text{h}}{\text{v}}\]
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Page 9
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Correct Answer: C
Solution :
\[\text{h=ut}-\frac{\text{1}}{\text{2}}\text{a}{{\text{t}}^{\text{2}}}\] \[\Rightarrow g{{t}^{2}}-2ut+2h=0\] Solving for t we get \[{{t}_{1}}+{{t}_{2}}=2u/g\] \[{{t}_{1}}\times {{t}_{2}}=2\,h\text{/}g\] \[\text{so, }\Delta \text{t= }\!\!|\!\!\text{ }{{t}_{1}}-{{t}_{2}}\text{ }\!\!|\!\!\text{ =}{{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-4{{t}_{1}}{{t}_{2}}\] Putting value we get \[\text{u=}\frac{1}{2}\sqrt{8hg+{{g}^{2}}\Delta {{t}^{2}}}\]
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Correct Answer: D
Solution :
Let the two balls P and Q meet at height x m from the ground after time t s from the start. We have to find distance, \[BC=\left( 100-x \right)\]For ball P \[\text{S=x m, u=25 m}{{\text{s}}^{-1}},\text{a}=-\text{g}\] From \[S=ut+\frac{1}{2}a{{t}^{2}}\] \[x=25t-\frac{1}{2}a{{t}^{2}}\] .......... (i) For ball Q \[\text{S= }\left( 100-\text{x} \right)\text{ m, u = 0, a =g}\] \[\therefore 100-x=0+\frac{1}{2}g{{t}^{2}}\] .......... (ii) Adding eqn. (i) and (ii), we get \[\text{100=25t or t=4s}\] From eqn. (i), \[\text{x=25}\times 4-\frac{1}{2}\times 9.8\times {{\left( 4 \right)}^{2}}=21.6\text{ m }\] Hence distance from the top of the tower \[\text{= (100-x)m=(100-21}\text{.6m)=78}\text{.4m}\]
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Correct Answer: B
Solution :
Let the body fall through the height of tower in t seconds. From, \[{{D}_{n}}=u+\frac{a}{2}\left( 2n-1 \right)\]we have, total distance travelled in last 2 second of fall is \[D={{D}_{t}}+{{D}_{\left( t-1 \right)}}\] \[=\left[ 0+\frac{g}{2}\left( 2t-1 \right) \right]+\left[ 0+\frac{g}{2}\left[ 2\left( t-1 \right) \right]-1 \right]\] \[=\frac{g}{2}\left( 2t-1 \right)+\frac{g}{2}\left( 2t-3 \right)=\frac{g}{2}\left( 4t-4 \right)\] \[=\frac{10}{2}\times 4\left( t-1 \right)\] \[\text{or, 40=20}\left( t-1 \right)\text{ or t=2+1=3s}\] Distance travelled at t second is \[\text{s=ut+}\frac{1}{2}a{{t}^{2}}=0+\frac{1}{2}\times 10\times {{3}^{2}}=45\text{ m}\]
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Page 14
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Correct Answer: C
Solution :
Given, \[H=10m\], \[u=5m/s,\text{ }g=10m/{{s}^{2}}\] Speed on reaching ground \[v=\sqrt{{{u}^{2}}+2gh}\] Now, v = u + at Now, \[v=\sqrt{{{u}^{2}}+2gh}=-u+gt\] Time taken to reach highest point is \[t=\frac{u}{g}\], \[\Rightarrow t=\frac{u+\sqrt{{{u}^{2}}+2gH}}{g}=\frac{nu}{g}\text{(from question)}\] \[\Rightarrow 2gH=n\left( n-2 \right){{u}^{2}}\] \[\Rightarrow m\left( n-1 \right)=\frac{2gH}{{{u}^{2}}}=\frac{2\times 10+10}{5\times 5}=8\Rightarrow n=4\]
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Page 15
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Correct Answer: A
Solution :
The distance travelled by the body A is A, given by \[{{v}_{1}}t=-\frac{g{{t}^{2}}}{2}\] and that travelled by the body B is \[{{\text{h}}_{2}}=\frac{g{{t}^{2}}}{2}\]s The distance between the bodies \[=x=h-({{h}_{1}}+{{h}_{2}}).\] Since\[{{\text{h}}_{\text{1}}}\text{+}{{\text{h}}_{\text{2}}}\text{=}{{\text{v}}_{\text{1}}}\text{t}\], the relation sought is \[x=h-{{v}_{1}}t\]
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Page 16
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Correct Answer: C
Solution :
\[\frac{\text{ }\!\!~\!\!\text{ Time of ascent}}{\text{Time of descent}}=\frac{\left( \frac{u}{g+a} \right)}{\frac{u}{\sqrt{\left( g+a \right)\left( g-a \right)}}}\] \[=\frac{\sqrt{\left( g+a \right)\left( g-a \right)}}{g+a}=\sqrt{\frac{\left( g-a \right)}{g+a}}\] \[=\sqrt{\frac{10-5}{10+5}}=\sqrt{\frac{5}{15}}=\sqrt{\frac{1}{3}}\]
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Page 17
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Correct Answer: D
Solution :
As the time taken from D to A = 2 sec. And D\[\to A\to B\to C=10\text{ sec}\left( given \right).\] As ball goes from \[B\to C\text{ }(\text{u=0, t=4sec})\] \[{{v}_{c}}=0+4g.\] As it moves from C to D, \[\text{s=ut+}\frac{1}{2}\text{a}{{\text{t}}^{2}}\] \[s=4g\times 2+\frac{1}{2}g\times 4=10g.\]
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Page 18
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Correct Answer: A
Solution :
\[t=\frac{s}{v}=\sqrt{{{H}^{2}}+{{L}^{2}}}\div \sqrt{\frac{2H}{g}}\]
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Page 19
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Correct Answer: B
Solution :
\[\text{v}_{\text{B}}^{\text{2}}\text{=v}_{\text{T}}^{\text{2}}\text{+2}\left( \text{10} \right)\left( \text{3} \right)\] \[\Rightarrow \text{v}_{\text{B}}^{\text{2}}\text{=v}_{\text{T}}^{\text{2}}\text{+60 }....\text{(i)}\] \[\text{Also, }{{\text{v}}_{B}}={{\text{v}}_{T}}+\left( 10 \right)\left( 0.5 \right)\text{ }....\text{(ii)}\] Solve eqs. (i) and (ii) We get \[{{\text{v}}_{T}}-{{\text{v}}_{B}}=4.9\text{ m/s}\]
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Page 20
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Correct Answer: B
Solution :
Ball A is thrown upwards from the building. During its downward journey when it comes back to the point of throw, its speed is equal to the speed of throw. So, for the journey of both the balls from point4to5. We can apply \[{{\text{v}}^{\text{2}}}-{{\text{u}}^{\text{2}}}\text{=2gh}\text{.}\] As u, g, h are same for both the balls, \[\,{{\text{v}}_{\text{A}}}\text{=}{{\text{v}}_{\text{B}}}\]
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