Physics

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

Let M be the mass of the rocket at any time t and v₁ the velocity of the rocket at the same time t

Let? m = mass of gas ejected in? t time

Relative speed of the gas ejected =u

KE +? t = KE of rocket +KE of gas

= ½ (M-? m) (v+? v)² + ½? m (v-u)²

KE_t= KE of rocket at time t= ½ Mv²

So? K = KE_t+? t -KE_t

= (M? v=? mu)v+1/2? mu²

Since action and reaction forces are equal

M? v/? t=? m/? t|u|

M? v=? m u

So? K= ½? mu²

? K=? W

? W=1/2? mu²

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

(a) As ball 1 is rolling down without slipping there is no dissipation of energy hence, total mechanical energy is conserved. Ball 3 is having negligible friction hence, there is no loss of energy.

(b) Ball 1 acquires rotational energy, ball 2 loses energy by friction. They cannot cross at C. Ball 3 can cross over.

(c) Ball 1, 2 turn back before reaching C. Because of loss of energy, ball 2 cannot reach back to A. Ball 1 has a rotational motion in “wrong” sense when it reaches B. It cannot roll back to A, because of kinetic friction.

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

(a) work done= increase in PE

= mg (vertical distance travelled)

= mg (s)sin $\theta =1*10*10sin30$ = 50J

(b) work done against friction = fs

= $\mu N\left(s\right)=\mu mgcos\theta$

= 0.1 $*1*10*cos30*10=8.66J$

(c) increase in PE =mgh

 =1 $*10*10*sin30$

$=100\left(\frac{1}{2}\right)=50J$

(d) according to work energy theorem W= change in KE

= -mgh-fs+FS

= -50-8.66+10 (10)

= 41.34J

(e) force f = FS

= 10 (10)= 100J

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) The length of the telescope tube L= f₀+f_e= 20+0.02= 20.02m

Also magnification is = 20/0.02= 1000

And image formed is inverted

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Morever, the formation of a virtual erect and enlarged image, takes place.

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) Alexandar's dark band lies between the primary and secondary rainbows, forms due to light Scattered into this region interfere destructively.

Since, primary rainbows subtends an angle nearly 41° to 42° at observer's eye, whereas, secondary rainbows subtends an angle nearly 51° to 54° at observer's eye w.r.t. incident light ray. So, the scattered rays with respect to the incident light of the sun lies between approximately 42°and50°.

This is a multiple choice answer as classified in NCERT Exemplar

(d) For µ = 16., the critical angle, µ = 1/ sin C, we have C = 38.7°, when viewed from AD, as long as angle of incidence on AD of the ray emanating from pin is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through AD.

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

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2x^2-3x-2}{x-2}\\ =\frac{2x^2-4x+x-2}{x-2}=\frac{2x\left(x-2\right)+1\left(x-2\right)}{x-2}\\ =\frac{\left(2x+1\right)\left(x-2\right)}{x-2}=2x+1\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2-h\right)+1=4+1=5\\ \underset{x\to 2^{+}}{lim}f\left(x\right)=2x+1=\underset{h\to 0}{lim}2\left(2+h\right)+1=4+1=5\\ \underset{x\to 2}{lim}f\left(x\right)=5\\ As\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)=5\\ Hence,f\left(x\right)iscontinuousatx=2.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.

The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.

This is a multiple choice answer as classified in NCERT Exemplar

(a) The negative refractive index metamaterials are those in which incident ray from air (Medium 1) to them refract or bend differently to that of positive refractive index medium.