Tags Physics NCERT Exemplar Solutions Class 11th Chapter Five

Physics NCERT Exemplar Solutions Class 11th Chapter Five

Get insights from 67 questions on Physics NCERT Exemplar Solutions Class 11th Chapter Five, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics NCERT Exemplar Solutions Class 11th Chapter Five

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m s^-2, what would be the reading of the weighing scale? (g = 10 m s^-2 )

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

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

Explanation – w^'=R=m (g-a)

Mass of person =50kg

Descending acceleration= 9m/s²

Acceleration due to gravity g= 10m/s²

Apparent weight of person R= m (g-a)

= 50 (10-9)=50N

Reading in weighing scale = 50/10 =5kg

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

A girl riding a bicycle along a straight road with a speed of 5 m s^-1 throws a stone of mass 0.5 kg which has a speed of 15 m ^s-1with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

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

explanation-total mass of bicycle and stone =m₁=50+0.5= 50.5kg

Velocity of bicycle u₁= 5m/s mass of stone m₂= 0.5kg

Velocity of stone u₂= 15m/s mass of girl and bicycle m= 50kg

Also the speed of bicycle change when stone is thrown

According to conservation of linear momentum

m₁u₁=m₂u₂+mv

50.5 (5)= 0.5 (15)+50 (v)

V= 4.9m/s

Change in speed 5-4.9= 0.1m/s

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

A body of mass 10kg is acted upon by two perpendicular forces, 6N and 8N. The resultant acceleration of the body is

(a) 1 m s^-2 at an angle of tan^-1(3/4) w.r.t. 6N force.

(b) 0.2 m s^-2 at an angle of tan^-1(3/4) w.r.t. 6N force.

(c) 1 m s^-2 at an angle of tan^-1(3/4) w.r.t.8N force.

(d) 0.2 m s^-2 at an angle of tan^-1(3/4) w.r.t.8N force.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – mass m= 10kg

F₁ =6N, F₂= 8N

Resultant force F= $\sqrt[]{36 + 64} = 10 N$

a=f/m= 10/10= 1m/s²

let $θ$ ₁ be the angle between R and F₁

tan $θ$ ₁=8/6=4/3

_{$θ$
1}= tan $θ$ ₁ (4/3) w.r.t F₁=6N

let $θ$ ₂ be the angle between F and F₂

tan $θ$ ₁=8/6=4/3

_{$θ$
2}= tan $θ$ ₂ (4/3) w.r.t F₂=8N

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of 5m s^-1 each, collide and rebound with the same speed. If the collision lasts for 10^-3 s, which of the following statements are true?

(a) The impulse imparted to each ball is 0.25 kg m s^-1 and the force on each ball is 250 N.

(b) The impulse imparted to each ball is 0.25 kg m s–1 and the force exerted on each ball is 25 * 10^-5 N.

(c) The impulse imparted to each ball is 0.5 Ns.

(d) The impulse and the force on each ball are equal in magnitude and opposite in direction.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- c, d

Explanation – m₁=m₂ = 50g= 1/20kg

Initial velocity u₁=u₂= 5m/s

Final velocity v₁=v₂= -5m/s

Time duration = 10^-3s

Change in linear ₁momentum = m (v-u)= 1/20 (-5-5)=-0.5N-s

Force = impulse /time=0.5/10^-3= 500N

So impulse and force are opposite in directions.

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

In Fig , a body A of mass m slides on plane inclined at angle θ1 to the horizontal and µ1 is the coefficient of friction between A and the plane. A is connected by a light string passing over a frictionless pulley to another body B, also of mass m, sliding on a frictionless plane inclined at angle θ2 to the horizontal. Which of the following statements are true?

(a) A will never move up the plane.

(b) A will just start moving up the plane when µ= $\frac{s i n θ_{2} - s i n θ_{1}}{c o s θ_{1}}$ .

(c) For A to move up the plane, θ₂ must always be greater than θ₁ .

(d) B will always slide down with constant speed.

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- b, c

Explanation- when A starts moving up

Mgsin $θ_{1} + f = m g s i n θ_{2}$

mgsin $θ_{1}$ + $μ m g c o s θ_{1}$ = $m g s i n θ_{2}$

$μ = \frac{s i n θ_{2} - s i n θ_{1}}{c o s θ_{1}}$

When A moves downward f= $m g s i n θ_{2} -$ mgsin $θ_{1}$ so clearly $θ_{2} > θ_{1}$

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

Mass m₁ moves on a slope making an angle θ with the horizontal and is attached to mass m₂ by a string passing over a frictionless pulley as shown in Fig. The co-efficient of friction between m₁ and the sloping surface is µ. Which of the following statements are true?

(a) If m₂ > m₁ sin θ , the body will move up the plane.

(b) If m₂ > m₁ (sin θ + µcos θ) , the body will move up the plane.

(c) If m₂ < m₁ (sin θ + µcos θ) , the body will move up the plane.

(d) If m₂ < m₁ (sin θ - µcos θ) , the body will move down the plane.

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- b, d

Explanation – as we know f = $μ N = μ m_{1} g c o s θ$

For system m₁+m₂ to move up

So m₂g- (m₁gsin $θ + f$ )>0

So m₂g- (m₁gsin $θ + μ m_{1} g c o s θ$ )>0

So m₂>m₁ (sin $θ + μ c o s θ$ )

But if bodies moves downward

m₁gsin $θ - f > m$ ₂g

m₁gsin> $θ + μ m_{1} g c o s θ$ m₂g

m₂1 (sinθ - uCosθ)

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

In Fig., the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shownon B. The mass of A is m/2 and of B is m. Which of the following statements are true?

(a) The bodies will move together if F = 0.25 mg.

(b) The body A will slip with respect to B if F = 0.5 mg.

(c) The bodies will move together if F = 0.5 mg.

(d) The bodies will be at rest if F = 0.1 mg.

(e) The maximum value of F for which the two bodies will move together is 0.45 mg.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- a, b, d, e

Explanation- suppose A and B are moving together a_common= $\frac{F - f_{1}}{m_{A} + m_{B}} = \frac{2 (F - t_{1})}{3 m}$

Pseudo force = m_A (a_common)= $\frac{m}{2} *$ $\frac{2 (F - t_{1})}{3 m}$ = $\frac{F - t_{1}}{3}$

Force will be maximum when pseudo force and frictional force are equal to one another

$\frac{F_{m a x} - f_{1}}{3}$ = $? m_{A} g$

= 0.2 $* \frac{m}{2} * g = 0.1 m g$

F_max= 0.3mg+f₁

= 0.3mg+0.1 (3/2)mg= 0.45mg

(a) for F=0.25mgmax bodies will move together

(b) for F=0.5mg>F_max body A will slip with respect to B

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin4 t for 0 < t<(1/4)s(A>0) and x = 0 for t >(1/4) s. Which of the following statements is true?

(a) The force at t = (1/8) s on the particle is –16π² A m.

(b) The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.

(c) The particle is not acted upon by any force.

(d) The particle is not acted upon by a constant force.

(e) There is no impulse acting on the particle.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- a, b, d

Explanation – x=0 for t<0s

X (t)=Asin4 $π t$ for 0

X=0, for t>1/4s

For,0 $π t$

Acceleration will be, a= dv/dt=-16 $π^{2}$ Asin4 $π t$

At t= 1/8 s, a (t)=-16 $π^{2}$ Asin4 $π (\frac{1}{8})$ =-16 $π^{2}$ A

So force F =ma =-16 $π^{2}$ Am

Impulse = change in linear momentum = f (t)= -16 $π^{2}$ Am (1/4)

Impulse = change in linear momentum

=F $* t$ =-16 $π^{2} A m * \frac{1}{4} = - 4 π^{2} A m$

= -4 $π^{2}$ Am so clearly force depends upon a so force is also not constant .

The impulse (change in linear momentum)

At t=0 is same as t=1/4s

Clearly, force depends upon A which is not constant. Hence, force is also not constant.

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

A car of mass m starts from rest and acquires a velocity along east v = vȋ (v>0) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is

(a) mv/2 eastward and is exerted by the car engine.

(b) mv/2 eastward and is due to the friction on the tyres exerted by the road.

(c) more than mv/2 eastward exerted due to the engine and overcomes the friction of the road.

(d) mv/2 exerted by the engine .

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- b

Explanation – mass of car m =0

Initial velocity =0,

velocity at east direction = v?

Time =2s

So v=u+at,

v? =0+a (2)

so a =v/2?

F=ma = $\frac{m v}{2} ?$ so force = mv/2 towards east.

0 0

New answer posted

3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Five Exemplar Solution

A body with mass 5 kg is acted upon by a force F= (–3ȋ + 4ĵ)N. If its initial velocity at t = 0 is v=(6ȋ-12ĵ) m/s , the time at which it will just have a velocity along the y-axis is

(a) Never

(b) 10 s

(c) 2 s

(d) 15 s

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions type Questions as classified in NCERT Exemplar

Answer- b

Explanation- we know that m= 5 kg

F= (–3? + 4? ) N and

initial velocity v= (6? -12? ) m/s

So retardation a= $\frac{F}{m}$ = ( $- \frac{3 ?}{5} + \frac{4 ?}{5}$ ) m/s²

We know v=u+at, for X-component only ,0 = 6? - $\frac{3 ?}{5} t$

So t= $\frac{5 * 6}{3}$ = 10s

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics NCERT Exemplar Solutions Class 11th Chapter Five

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience