physics ncert solutions class 11th

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

When speed becomes constant then acceleration will be zero

a=dv/dt=0

But here a=g-bv

Clearly from above equation as speed increases acceleration will decrease . at a certain speed say v_o, acceleration will be zero and speed will remain constant.

Hence So g=bv

so v= g/b

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

x (t)=A+B

Let A>B and

Now velocity is equal to x (t)=dx/dt=-B

So a (t)= dv/dt= B

Above condition are satisfied by the equation.

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

(a) The equation we use here is x= 1-sint

So velocity = dx/dt=1-cost

Acceleration =sint

When t=o x=o

When t=, x=

When t=0.x= 2

(b) x=sint so velocity becomes v=cost as displacement and velocity contain sin and cos so equation is periodic.

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

If effect of gravity is neglected then ball moving uniformly turned back with same speed when a ball hit it. Acceleration of the ball is zero just before it hits the bat and due  to the repulsive force it gets accelerated.

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

We have to analyse slope of each curve i.e= dx/dt . for peak values dx/dt will be zero as x is maximum at peak points.

For graph (a) there is appoint for which displacement is zero so a matches with (iii)

In graph b, x is positive throughout and at point B, V=dx/dt=0

Since at point of curvature changes a=0, so b matches with (ii)

displacement is zero in only first graph so it matches with the (iii) point.

And slope of d graph v=dx/dt is positive so v>0 so acceleration will be negative so matches with I but in graph c it matches with iv as its slope is negative.

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

let speed of two balls be V₁and V₂

Where v₁=2v and v₂=v and y₁and y₂ be the distance covered

So y₁= $\frac{v_1^2}{2g}=\frac{4v^2}{2g}$ and y₂= $\frac{v_2^2}{2g}=\frac{v^2}{2g}$

So y₁-y₂= 15

$\frac{3v^2}{2g}=15$

V²= $\sqrt[]{5*2*10}=\frac{10m}{s}$

So clearly we can say v₁=20 and v₂=10

And y₁=20m and y₂=5m

If t₂ is the time taken by ball 2 through a distance of 5m, y₂=v₂t-1/2gt²

5=10t₂-5t₂² so t₂ will be 15

Then time covered by ball 1 in 2 sec between two throws = t₁-t₂= 2-1=1s

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

(a)  for maximum velocity dv/dt=0

d/dt (6t-2t²)=0

6-4t=0 t= 6/4=1.5s

(b) v=6t-2t²

ds/dt=6t-2t²

ds=6t-2t²dt

distance in 3s, S= ${\int }_0^3\left(6t-{2t}^2\right)dt=[3t^2-\frac{2}{3}{t}^3]$ ³₀

s= 27-18=9m

average velocity = distance /time =9/3 = 3m/s

x= 6t-2t²

3=6t-2t²

After solving we get t= 9/4s approx.

(c) in periodic motion when velocity is zero

0=6t-2t²

0=t (6-2t)

So t=0, 3 sec

(d) distance covered from 0 to 3s=9m

distance covered in 3 to 6s= ${\int }_3^6\left(18-9t+t^2\right)dt$

S= (18t- $\frac{9t^2}{2}+\frac{t^3}{3}$ )⁶

S= 108-9 (18)+ $\frac{6^3}{3}-18\left(3\right)+\frac{9}{2}\left(9\right)-\frac{27}{3}$

S= -4.5m

So total distance covered = 9+ (-4.5)=4.5m

No of cycles covered in that distance =20/4.5=4.44approx

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

speed of car and truck = 72km/h = 72 (5/18) =20m/s

V= u+at

0=20+a (5) so a=-4m/s²

But retarted acceleration will be v=u+at

0=20+a (3)

So a= $\frac{20}{3}\left(t-0.5\right)$ -20/3m/s²

We also need to consider human response time = 0.5 s

V=u-at (for retarded motion)

V= 20- $\frac{20}{3}\left(t-0.5\right)$ ….1

V_t=20-4t ….2

From 1 and 2

20-=20-4t

After solving we get t= 5/4s

Distance travelled by truck in time t, S=ut+1/2at²

= 20 $*\frac{5}{4}+\frac{1}{2}\left(-4\right)*{\left(\frac{5}{4}\right)}^2=21.875m$

To avoid the bump onto the truck car must maintain distance = 23.125-21.875=1.250m

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

(a) velocity attained by a falling rain drop will be = $\sqrt[]{2gh}=\sqrt[]{2*10*1000}$

= $100\sqrt[]{2}m{s}^{-1}=\frac{510km}{h}$

(b) diameter of the rain drop = 2r=4mm

Radius = 2mm= 2 $*{10}^{-3}m$

Mass of rain drop = V $*\rho =\frac{4}{3}\pi r^3\rho =\frac{4}{3}*\frac{22}{7}*{\left(2*{10}^{-3}\right)}^3*{10}^3=3.4*{10}^{-5}kg$

Momentum of rain drop= mv= 3.4 $*{10}^5*100\sqrt[]{2}=5*{10}^{-3}kgm/s$

(c) time ,t = d/v= $\frac{4*{10}^{-3}}{100\sqrt[]{2}}=0.028*{10}^{-3}s$

(d) force exerted, F = change in momentum /time=

(e) area = $\pi R^2=\pi *{\frac{1}{2}}^2=\frac{11}{14}=0.8m^2$

number of drops striking the the umbrella with separation of 5 $*{10}^{-2}$

so net force = $\frac{0.8}{{(5*{10}^{-2})}^2}*168=53760N$