physics ncert solutions class 11th

This is a long answer type question as classified in NCERT Exemplar

Let us assume t=0 when $\theta =\theta$ _o then $\theta$ = $\theta$ ₀coswt

Given a seconds pendulum w=2 $\pi$

$\theta$ = ${\theta }_{o}cos2\pi t$

At time t, let $\theta ={\theta }_o/2$

Cos2 $\pi t_1=\frac{1}{2}$  , t₁=1/6

$\frac{d\theta }{dt}$ =-( ${\theta }_{o}2\pi$ )sin2 $\pi t$

t=t₁=1/6

d $\frac{\theta }{dt}$ =- ${\theta }_{o}2\pi$ sin $\frac{2\pi }{6}=-\sqrt[]{3}\pi {\theta }_o$

the linear velocity is u=- $\sqrt[]{3}\pi {\theta }_{o}l$

the vertical component is U_y= $-\sqrt[]{3}\pi lsin\left(\frac{{\theta }_o}{2}\right)$

the horizontal component U_x=- $-\sqrt[]{3}\pi lcos\left(\frac{{\theta }_o}{2}\right)$

at the time it snaps the vertical height is

H'=H+l(1-cos $\left(\frac{{\theta }_o}{2}\right)$ )

Let the time require for fall be t , then

H'= H+l (1 $-cos\left(\frac{{\theta }_o}{2}\right)$ )

Let the time required for fall be at t then

H'=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3}\pi {\theta }_{o}lsin\frac{{\theta }_o}{2}t-H^{\text{'}}=0$

t= $\frac{-\sqrt[]{3}\pi {\theta }_{0}sin\frac{{\theta }_o}{2}\pm \sqrt[]{3{\pi }^2{\theta }_o^2{sin}^2\frac{{\theta }_o}{2}+2gH\text{'}}}{g}$

given that ${\theta }_o$ is small , hence neglecting terms of order ${\theta }_o^2$ and higher

t= $\sqrt[]{\frac{2H^{\text{'}}}{g}}$

H' $\approx H+l(1-1)$

t= $\sqrt[]{\frac{2H}{g}}$

the distance travelled in the x direction is u_xt to the left of where the bob is snapped 

X= Uxt= $\sqrt[]{3}\pi {\theta }_{o}lcos\left(\frac{{\theta }_o}{2}\right)\sqrt[]{\frac{2H}{g}}s$

cos $\frac{{\theta }_o}{2}\approx 1$

X= $\sqrt[]{3}\pi {\theta }_{o}l\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

At the time of snapping , the bob was at a horizontal distance of $lsin\frac{{\theta }_o}{2}\approx l\frac{{\theta }_0}{2}$ from A

Thus, the distance of bob from A where it meets the ground is

= $\frac{l{\theta }_o}{2}-X=\frac{l{\theta }_o}{2}-\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi \right)$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}\right)$

This is a long answer type question as classified in NCERT Exemplar

(a) When the support of the hand is removed the body oscillates about mean position

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.

Loss in PE of the block=mgx

Gain in elastic potential energy =1/2 kx²

By energy conservation we cam say that

Mgx=1/2kx²

Or x= 2mg/k

Now the mean position of oscillation will be when the block is balanced by spring

If x' is the extension in that case

F= kx'

F=mg

Mg=kx'

X'=mg/k

By dividing x by x'

x/x'= $\frac{2mg/k}{mg/k}=2$

so x=2x'

x'=4/2 =2cm

but the displacement of mass from the mean position when spring attains its natural length is equal to amplitude of the oscillation

A=x'=2cm

(b) Time period of the oscillating system depends on mass spring constant given by

T= $2\pi \sqrt[]{\frac{m}{k}}$

2mg/k=x

2mg/k=4cm =4 $*{10}^{-2}m$

m/k =4 $*{10}^{-2}$ /2g

k/m=g/2 $*{10}^{-2}$

$v=\frac{1}{T}\sqrt[]{\frac{k}{m}}$

$v=\frac{1}{2*3.14}\sqrt[]{\frac{g}{2*{10}^{-2}}}$

= $\frac{10}{628}*2.21$ =3.51Hz

This is a long answer type question as classified in NCERT Exemplar

Let us assume t=0 when $\theta =\theta$ _o then $\theta$ = $\theta$ ₀coswt

Given a seconds pendulum w=2 $\pi$

$\theta$ = ${\theta }_{o}cos2\pi t$

At time t, let $\theta ={\theta }_o/2$

Cos2 $\pi t_1=\frac{1}{2}$  , t₁=1/6

$\frac{d\theta }{dt}$ =-( ${\theta }_{o}2\pi$ )sin2 $\pi t$

t=t₁=1/6

d $\frac{\theta }{dt}$ =- ${\theta }_{o}2\pi$ sin $\frac{2\pi }{6}=-\sqrt[]{3}\pi {\theta }_o$

the linear velocity is u=- $\sqrt[]{3}\pi {\theta }_{o}l$

the vertical component is U_y= $-\sqrt[]{3}\pi lsin\left(\frac{{\theta }_o}{2}\right)$

the horizontal component U_x=- $-\sqrt[]{3}\pi lcos\left(\frac{{\theta }_o}{2}\right)$

at the time it snaps the vertical height is

H'=H+l(1-cos $\left(\frac{{\theta }_o}{2}\right)$ )

Let the time require for fall be t , then

H'= H+l (1 $-cos\left(\frac{{\theta }_o}{2}\right)$ )

Let the time required for fall be at t then

H'=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3}\pi {\theta }_{o}lsin\frac{{\theta }_o}{2}t-H^{\text{'}}=0$

t= $\frac{-\sqrt[]{3}\pi {\theta }_{0}sin\frac{{\theta }_o}{2}\pm \sqrt[]{3{\pi }^2{\theta }_o^2{sin}^2\frac{{\theta }_o}{2}+2gH\text{'}}}{g}$

given that ${\theta }_o$ is small , hence neglecting terms of order ${\theta }_o^2$ and higher

t= $\sqrt[]{\frac{2H^{\text{'}}}{g}}$

H' $\approx H+l(1-1)$

t= $\sqrt[]{\frac{2H}{g}}$

the distance travelled in the x direction is u_xt to the left of where the bob is snapped 

X= Uxt= $\sqrt[]{3}\pi {\theta }_{o}lcos\left(\frac{{\theta }_o}{2}\right)\sqrt[]{\frac{2H}{g}}s$

cos $\frac{{\theta }_o}{2}\approx 1$

X= $\sqrt[]{3}\pi {\theta }_{o}l\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

At the time of snapping , the bob was at a horizontal distance of $lsin\frac{{\theta }_o}{2}\approx l\frac{{\theta }_0}{2}$ from A

Thus, the distance of bob from A where it meets the ground is

= $\frac{l{\theta }_o}{2}-X=\frac{l{\theta }_o}{2}-\sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}{\theta }_{o}l\pi \right)$

= ${\theta }_{o}l\left(\frac{1}{2}-\pi \sqrt[]{\frac{6H}{g}}\right)$