Class 11th

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

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) We know that one light year = 9.46 $*{10}^{11}m$ = distance that light travels in 1 year with a speed 3 $*{10}^8$ m/s

1 par sec =3.08 $*{10}^{16}m$ = distance at which average radius of earth's orbit subtends an angle of par second.

Here second and year represent time.

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) we know that pressure = force/area

Pressure= $\frac{force*distance}{area*distance}=\frac{work}{volume}=\frac{energy}{volume}$

So we can say that pressure is also equal to energy per unit volume.

This is a multiple choice answer as classified in NCERT Exemplar

(a,b,d) h=[ML²T^-1]

[c]=[LT^-1], [m_e]=M

[G]= [M^-1L³T^-2]

[e]=[AT],[m_p]=[M]

[ $\frac{hc}{G}$ ] = $\frac{\left[M{L}^2{T}^{-1}\right]\left[L{T}^{-1}\right]}{M^{-1}{L}^3{T}^{-2}}=M^2$

M= $\sqrt[]{\frac{hc}{G}}$

$\frac{h}{c}=\frac{M{L}^2{T}^{-1}}{L{T}^{-1}}=\left[ML\right]$

L= $\frac{h}{cM}=\frac{h}{c}\sqrt[]{\frac{G}{hc}}=\frac{\sqrt[]{Gh}}{c^{3/2}}$

C= LT^-1

T= $\frac{\left[L\right]}{\left[C\right]}$ = $\frac{\sqrt[]{Gh}}{c^{3/2}c}=\frac{\sqrt[]{Gh}}{c^{5/2}}$

Hence a,b and d any can be used to express L,M and T in terms of three chosen fundamental quantities.

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) as we know E= hv

So h=E/v= $\frac{ML{T}^{-2}}{T^{-1}}$ =ML²T^-1

Dimension of linear impulse = dimensions of momentum= [MLT^-1]

As we know that J= $?p$

Angular impulse =tdt= $?L$ = change in angular momentum

Hence angular momentum is [ML²T^-1] same as the dimension of h.

This is a multiple choice answer as classified in NCERT Exemplar

(a, e) It is given that P, Q, R are having different dimensions. Hence cannot be added or subtracted, so we can say that (a) and (e) are not meaningful. We cannot say about the dimensions of product of these quantities. Hence b, c and d may be meaningful.

This is a multiple choice answer as classified in NCERT Exemplar

(b, c) for (c) LHS = [L]

RHS= [L/T]=LT^-1

LHS $\ne$ RHS

Hence is c option is not correct.

In option (b) dimension of angle is vt i.e =L

RHS= L.L=L² and LHS = L

LHS $\ne$ RHS

So option b is also not correct

LHS and RHS in both equation B and C are not equal.