Physics NCERT Exemplar Solutions Class 11th Chapter Four

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- if the football is kicked into the air vertically upwards. Acceleration of football will always vertically downwards and equal to acceleration due to gravity. But when it reaches the highest point its velocity is zero and acceleration is retarding.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- at point B it will gain the same speed u and after that speed increases and will be maximum just before reaching point c.

During journey from O to A speed decreases and will be maximum at point A and acceleration is always constant.

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Explanation- vsin $\theta$ = vertical component

V_x= horizontal component of velocity =vcos $\theta$ = constant

V_y = vertical component of velocity =vsin $\theta$

Velocity will always be tangential to the curve in the direction of motion and acceleration is always vertically downward and is equal to g.

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – T_sand= $\frac{AP+QC}{1}+\frac{PQ}{v}=\frac{25\sqrt[]{2}+25\sqrt[]{2}}{1}+\frac{50\sqrt[]{2}}{v}$

 = 50 $\sqrt[]{2}+\frac{50\sqrt[]{2}}{v}=50\sqrt[]{2}\left(\frac{1}{v}+1\right)$

Time taken T_outside= $\frac{AR+RC}{1}s$

AR= $\sqrt[]{{75}^2+{25}^2}=25\sqrt[]{10}m$

RC= AR= $25\sqrt[]{10}m$

T_outside= 2AR= 50 $\sqrt[]{10}s$

T_sandoutside  . so After solving we get velocity is greater v= 0.81m/s

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Explanation – r=cos $\theta ?+sin\theta$ ?…….1

$\theta =-sin\theta ?+cos\theta ?$ …….2

Multiplying eq1 by sin $\theta$  and 2 with cos $\theta$ and adding

Rsin $\theta +\theta cos\theta =sin\theta cos\theta ?+{sin}^2\theta j+{cos}^2\theta ?-sin\theta cos\theta ?$

= ?( ${cos}^2\theta +{sin}^2\theta$ )=j

= rsin $\theta +\theta cos\theta =j$

 n(rcos $\theta -\theta sin\theta$ )=i

b)r $\theta =cos\theta ?+sin\theta ?(-sin\theta ?+cos\theta ?)$

 = -cos $\theta sin\theta +sin\theta .cos\theta =0$ $\theta =90$

c)r=cos $\theta ?+sin\theta ?$

dr/dt=d/dt(cos $\theta ?+sin\theta ?$ )=w[-cos $\theta ?+sin\theta ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $\theta r=\theta [\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= dr/dt= $\frac{d\theta }{dt}r+\theta \frac{d}{dt}[\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]$

v= $\frac{d\theta }{dr}r+\theta [-\mathrm{c}\mathrm{o}\mathrm{s}\theta ?+sin\theta ?]\frac{d\theta }{dt}$

= $\frac{d\theta }{dt}r+\theta w\theta =wr+$ w $\theta \theta$

a= $\frac{d}{dt}\left[wr+w\theta \theta \right]=\frac{d}{dt}\left[\frac{d\theta }{dt}r+\frac{d\theta }{dt}\left(\theta \theta \right)\right]$

a= $\frac{d^2\theta }{d{t}^2}r+\frac{d\theta }{dt}$ $\frac{dr}{dt}+\frac{d^2\theta }{d{t}^2}\theta \theta +\frac{d\theta }{dt}\frac{d}{dt}\left(\theta \theta \right)$

= $\frac{d^2\theta }{d{t}^2}r+w^2\theta +\frac{d^2\theta }{d{t}^2}\theta \theta +w^2\theta +w^2\theta (-r)$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- a) for x direction u_x= u+v_ocos $\theta$

u_y=velocity in y direction= v₀sin $\theta$

now tan $\theta =\frac{u_y}{u_x}=\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

$\theta ={tan}^{-1}\frac{u_{o}sin\theta }{u+u_{o}cos\theta }$

b) let t be the time flight y =0 u_y=v_osin $\theta ,a_y=-g,t=T$

y= u_yt+1/2 a_yt²

0= v_osin $\theta T$ + $\frac{1}{2}\left(-g{T}^2\right)$

So T = $\frac{2u_{o}sin\theta }{g}$

c) horizontal range R, = (u+v_ocos $\theta$ T= (u+v_ocos $\theta$ ) $\frac{2u_{o}sin\theta }{g}$

d) for range to be maximum dR/d $\theta =0$

$\frac{v_o}{g}\left[2ucos\theta +v_{o}cos2\theta *2\right]=0$

$2ucos\theta +2v_o\left[2{cos}^2\theta -1\right]=0$

4v_ocos^{2 $\theta +2ucos\theta -2v_o=0$}

So cos $\theta$ = $\frac{-u?\sqrt[]{u^2+8v_o^2}}{4v_o}$

$\theta ={cos}^{-1}\left[\frac{-u\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]$

e) cos $\theta$ = $\frac{-v_o?\sqrt[]{u^2+8v_o^2}}{4v_o}=\frac{-1+3}{4}=\frac{1}{2}$

so $\theta =60$

f) if u=0 ${\theta }_{max}={cos}^{-1}\left[\frac{-0\pm \sqrt[]{u^2+8v_o^2}}{4v_o}\right]={cos}^{-1}\left(\frac{1}{\sqrt[]{2}}\right)=45$ ⁰

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_r^2+v_s^2}=\sqrt[]{3^2+4^2}=5m/s$

tan so 'N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_s^2-v_r^2}=\sqrt[]{4^2-3^2}=\sqrt[]{7}m/s$

tan $\theta =\frac{v_r}{v}=\frac{3}{\sqrt[]{7}}$

(c) time taken by swimmer t =d/v= d/4s

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1 

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II 

v_g = 10m/s?

V_rg= V_r - V_g

 = a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =1

So b =-5

Velocity of rain = a? +b?

V_r = 5? -5?

Speed of rain V_r= $\sqrt[]{5^2+{(-5)}^2}=5\sqrt[]{2}m/s$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- y=O, u_y= V_ocos $\theta$

a_y=-gcos $\theta$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2}{a}_y$ t²

0 = V_ocos $\theta T$ +T^{2 $\frac{1}{2}(-gcos\theta )$}

T= $\frac{2V_{o}cos\theta }{gcos\theta }$

T= 2V₀/g

X= L, u_x=V_osin $\theta$  , a_x= gsin $\theta$  , t=T= $\frac{2V_o}{g}$

X=u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V_osin $\theta T+\frac{1}{2}gsin\theta T^2$

L= $\frac{4v_o^2}{g}$ sin $\theta$