JEE SECTION - A [PHYSICS MCQ]

$-H =4 \times 4-\frac{1}{2} \times 10 \times 4^2$

$-H =16-80$

$-H =-64$

$H =64\,m$

$\left(\frac{u}{3}\right)^2=u^2-2 a x$

$2 a x=u^2-\frac{u^2}{9}$

$2 a x=\frac{8 u^2}{9}.........(1)$

Similarly from starting

$v^2=u^2+2 ax$

$0= u ^2-2 ax _2$

$2 ax _2= u ^2..........(2)$

$By (1) /(2)$

$\frac{ x }{ x _2}=\frac{8}{9}$

$\frac{24}{ x _2}=\frac{8}{9}$

$x _2=27\,cm$

$=0+\frac{ a }{2}(2 n -1)$

$S _{r th } \propto(2 n -1)$

$\Rightarrow S_{1 s }, S _{2 ed }, S _{3 ed }, S _{\text {4th }}$

$=[2(1)-1]:[2(2)-1]:[2(3)-1]:[2(4)-1]$

$=1: 3: 5: 7$

$V =\frac{ dx }{ dt }=\tan \theta$

$\frac{ V _{1}}{ V _{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}}=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}=\frac{1}{\sqrt{3}}$

$S = ut +\frac{1}{2} at ^{2}$

$1.5= u (0.1)+\frac{1}{2} \times 10(0.1)^{2}$

$1.5=(0.1) u +0.05$

$u =15-0.5$

$\quad=14.5 m / s$

$80^{2}=20^{2}+2 \times 10 h$

$h =300 m$

$\begin{array}{l}
\therefore \,Net\,velocity\,of\,preeti\,on\,moving\,escalator\,with\,respect\,\\
to\,the\,ground\,\\
\,\,\,\,\,\,\,\,\,\,\,v = {v_1} + {v_2} = \frac{d}{{{t_1}}} + \frac{d}{{{t_2}}} = d\left( {\frac{{{t_1} + {t_2}}}{{{t_t}{t_2}}}} \right)\\
The\,time taken by her to walk up on the moving escalatror will be\\
\,\,\,\,\,\,t = \frac{d}{v} = \frac{d}{{d\left( {\frac{{{t_1} + {t_2}}}{{{t_1}{t_2}}}} \right)}} = \frac{{{t_1}{t_2}}}{{{t_1} + {t_2}}}
\end{array}$

$\frac{ ds }{ dt }=\alpha t +\beta t ^{2}$

$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$

$S_{2}-S_{1}=\left[\frac{\alpha t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{1}^{2}$

As particle is not changing direction So distance $=$ displacement.

Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$

$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$

$s_{ rel }=u_{ rel } t+\frac{1}{2} a_{ rel } t^2$

$a_{ rel }=0$

$\Rightarrow 40=(10-6) \times t$

$\Rightarrow \frac{40}{4}=t \Rightarrow t=10 \,s$

When two trains are moving in opposite direction then

$v_{\text {rel }}=(20+10)=30 \,ms ^{-1}$

$t=\frac{200}{30}=6.67 \,s$

$v^2=u^2+2 a d$

$\Rightarrow 0=(10)^2-2 \times 2 \times d_{ rel }$

$\Rightarrow \frac{100}{4} \leq d_{ rel }$

$\Rightarrow d_{ rel } \geq 25 \,m$

The distance of $5 km$ is in between $A$ and $B$ is covered by $C$ in 4 minute with relative velocity $(v+30)$.

So, $d_{\text {rel }}=v_{\text {rel }} \times t$

$\Rightarrow 5 \,km =(v+30) \times \frac{4}{60}$

$\Rightarrow 75 \,kmh ^{-1}=v+30$

$\Rightarrow v=45 \,kmh ^{-1}$

$S_{ rel }=U_{ rel } t+\frac{1}{2} a_{ rel } t^2$

$\Rightarrow S_{ rel }=(60-40) 5 \quad\left(a_{ rel }=0\right)$

$\Rightarrow S_{ rel }=100 \,m$

Given that:

velocity of blocks $=10\,ms ^{-1}$

Friction force between $B$ and $A=20\,N$.

As we know,

Force $=$ mass $\times$ acceloration

So, Friction force between the block $B$ and the ground is zero as acceleration of the blocks is zero as they are moving with constant velocity.

Relative to lift

$u_r =0$

$a_r =(9.8+1.2)=11\,m / s ^2$

$s_r =\frac{1}{2} a_r t^2$

$2.7 =\frac{1}{2} \times 11 \times t^2$

or $t=\sqrt{\frac{5.4}{11}}\,s$

$60 \mathrm{km}$ to $\mathrm{m} / \mathrm{s} \rightarrow\left(60 \times \frac{5}{18}\right)$

$V_{A B}=\left(V_{A}-V_{B}\right)=\frac{5}{18}[70+60]=\left(\frac{5}{18} \times 130\right)$

$a_{A B}=\left(a_{A}-a_{B}\right)=-(5+5)=-10 m / s^{2}$

$v^{2}=u^{2}+2 a s$

$o=u^{2}-2 \times 80 \times 10$

$u^{2} \leqslant 800 \times 2=1600$ (To revert collision)

$1304.0 \leqslant 1600$

$v_1=\frac{s}{90}, v_2=\frac{s}{60}$

Now,

$t=\frac{s}{v_1+v_2}=\frac{s}{\frac{s}{90}+\frac{s}{60}}$

$=\frac{90 \times 60}{90+60}=36\,s$

$7.2=\frac{100}{(v+5)(5 / 18)}$

or

$v=45\,km / h$

$\therefore \frac{x-u}{x+u}=\frac{1}{2}$ or $2 x-2 u=x+u$

or $x=3 u$

$v_1+v_2=8 / 1=8$

In same direction they subtract

$v_1-v_2=0.8 / 1=0.8$

From both eqn,

${v}_1=4.4$

${v}_2=3.6$

$v_A+v_B=4\,m / s$ and $v_A-v_B=\frac{4.0}{10}=0.4\,m / s$

Slope of $B$ is more. Therefore, speed of $B$ will be more.

$S\, = \,{V_{rel.}}\, \times \,t\,\, = \,81\,\, \times \,\frac{5}{{60}}\, = \frac{{81}}{{12}} = 6.75\,\,\,km$

Displacements of both should be equal. Or,

$8 t=\frac{1}{2} \times 4 \times t^2$ or $t=4\,s$

Distance travelled by bird in this time

$s=v_{3} t=\frac{v_{3} d}{v_{1}+v_{2}}=\frac{10 \times 2000}{(20+30)}=400 m$

$t=\frac{10 \cos 45}{10 \sqrt{2}}$

$v+30=\frac{5}{\left(\frac{4}{60}\right)}$

$\Rightarrow v+30=\frac{5 \times 60}{4}=75$

$\Rightarrow v=75-30=45 \mathrm{km} / \mathrm{h}$

$1.5 m / s^{2}$

because time taken to stop wrt belt is $4\; sec$. Distance travelled wrt belt is

$S=6 \times 4-\frac{1}{2} \times 1.5 \times 4^{2}=12$

Distance travelled by belt in that time is

$4 \times 4=16$

Now taking direction of movement of belt as positive . Displacement wrt to belt is - $12 \mathrm{m}$. Displacement wrt ground is $(-12+16=4)=4 \mathrm{m}$

Also displacement wrt ground will be zero when

$S=6 \times t-\frac{1}{2} \times 1.5 \times t^{2}-4 \times t=0$

$\Rightarrow t=\frac{8}{3}$

acceleration due to gravity $a_{g}=-g$ relative acceleration

$a_{r e l}=-g-a$ $t_{r e l}=\frac{2 u}{g+a}$

$a=\frac{2 u-g t}{t}$

$=10+10=20m/s.$

Relative acceleration $=0.3+0.2=0.5 m/s^2$

If trains cross each other then from $s = ut + \frac{1}{2}a{t^2}$

$As,\;s = {s_1} + {s_2}$$ = 100 + 125 = 225$

==> $225 = 20t + \frac{1}{2} \times 0.5 \times {t^2}$ ==> $0.5{t^2} + 40t - 450 = 0$

==> $t = - \frac{{40 \pm \sqrt {1600 + 4.(005) \times 450} }}{1}$$ = - 40 \pm 50$

$\therefore t = 10\sec $ (Taking +ve value).

From $v = u - at$ $⇒$ $0 = ({v_1} - {v_2}) - at$ $⇒$ $t = \frac{{{v_1} - {v_2}}}{a}$

$\therefore $Time for trip to that side $ = \frac{{1km}}{{4km/hr}} = 0.25\,hr$

To come back, again he take $0.25$ hr to cross the river.
Total time is $30\, min$, he goes to the other bank and come back at the same point.

When these particles moves in the same direction then ${v_1} - {v_2} = 4$...(ii)

By solving ${v_1} = 5$ and ${v_2} = 1\;m/s$

$\therefore  t = \frac{{150}}{{15}} = 10\;sec$

$ = 180\;m/s + 45\;km/h = (180 + 12.5)\;m/s = 192.5\;m/s$

Speed of thief ’s jeep $ = 153\,km/h = 42.5m/s$

Velocity of bullet w.r.t thief ’s car $ = 192.5 - 42.5$

$=150\,m/s$

Relative velocity $ = 30 - ( - 20) = 50\;m/s$

Hence $t = 250/50 = 5s$

$v_A=10 \,ms ^{-1}-10 t$

$v_B=0-10 t$

$v_{A B}=v_A-v_B=10-(10 t)-(-10 t)=10-10 t+10 t=0$

$\Rightarrow v_{A B}=10 \,ms ^{-1}$

$u_{\text {rel }}=u_1-u_2=0-(-5)=5 ms ^{-1}$

$t=3 \,s$

$a_{ rel }=a_1-a_2=-g-(-g)=0 \,ms ^{-2}$

$s_{ rel }=u_{ rel } t+\frac{1}{2} a_{ rel } t^2$

$\Rightarrow s_{ rel }=5 \times 3=15 \,m \quad\left(\because a_{ rel }=0\right)$

So, $s_{\text {rel }}=15 \,m$

Let the total height of tower $=H$

Total time of journey $=t$

Time taken to cover the $\frac{5 h}{9}$ is = last second

So, $s_t-s_{t-1}=\frac{5 h}{9}$

$\Rightarrow \frac{1}{2} g t^2-\frac{1}{2} g(t-1)^2=\frac{5}{9} \times \frac{1}{2} g t^2$ $\left[\because h=\frac{1}{2} g t^2\right]$

$\Rightarrow \frac{1}{2} g\left(t^2-t^2-1+2 t\right)=\frac{1}{2} g t^2 \times \frac{5}{9}$

$\Rightarrow (2 t-1)=\frac{5}{9} t^2$

$\Rightarrow 18 t-9=5 t^2$

$\Rightarrow 5 t^2-18 t+9=0$

$\Rightarrow 5 t^2-15 t-3 t+9=0$

$\Rightarrow 5 t(t-3)-3(t-3)=0$

$\Rightarrow(5 t-3)(t-3)=0$

$t=\frac{3}{5}, t=3 s \quad\left(t=\frac{3}{5}\right.$, doesn't satisfy the given criterion, so we neglect it)

$v^2-u^2=-2 g \times \frac{2 H}{3}$

$\Rightarrow-100 \times 2=-2 \times 10 \times \frac{2 H}{3}$

$\Rightarrow H=15 \,m$

If the ball is dropped then $x=0$, the velocity with which it will hit the sand will be given by

$v^2-u^2=2(-g)(-9)$

$v^2-0=18 g$

$v^2=18 g$

Now on striking sand, the body penetrates into sand for $1 m$ and comes to rest. So, $v \rightarrow$ initial for sand and final velocity $=0$

$v^{\prime 2}-v^2=2(a) \times(-1)$

$\Rightarrow -18 g=-2 a$

$\Rightarrow a=9 g$

$\frac{-H}{2}=u t-\frac{1}{2} g \times 10^2$

$\Rightarrow H=g \times 10^2$

$\Rightarrow -H=-\frac{1}{2} g t^2 \quad$ (Full joumey)

$g \times 10^2=\frac{1}{2} g t^2$

$\Rightarrow t^2=200$

$\Rightarrow t=10 \sqrt{2} s$

$\Rightarrow t=10 \times 1.414 s$

$=14.14 s =t$

$v=\sqrt{2 g(h-y)}$