## A particle moving on straight line whose velocity-time graph is shown in the figure. The average speed from t = 0 to t = 6 s is …

Q: A particle moving on straight line whose velocity-time graph is shown in the figure. The average speed from t = 0 to t = 6 s is v = 15/n m/s. Find the value of n. Solution : Area of v-t graph from zero to 5 sec is

A1 = Area of trapezium

$\displaystyle A_1 = \frac{1}{2}(5+3) \times 10$

$\displaystyle A_1 = 40 m$

Area of v-t graph from 5 sec to 6 sec is

A2 = Area of triangle

$\displaystyle A_2 = \frac{1}{2} \times 1 \times 10$

A2 = 5 m

Total distance traveled = 40 + 5 = 45 m

Average Speed = Total Distance / Total time

$\displaystyle V = \frac{45}{6} = \frac{15}{2}$

On comparing we get ,

n = 2

## A particle is moving on a straight line whose velocity as function of time is (t-2) m/s . Find the distance traveled …

Q: A particle is moving on a straight line whose velocity as function of time is (t-2) ms-1. Find distance travelled by particle (in m) in 4 s.

Solution : Here , v= 0 at t = 2 sec

Distance travelled by particle in 4 sec is

$\displaystyle S = |\int_{0}^{2} v dt | + |\int_{2}^{4} v dt |$

$\displaystyle S = |\int_{0}^{2} (t-2) dt | + |\int_{2}^{4} (t-2) dt |$

$\displaystyle S = |[\frac{t^2}{2} – 2 t]_{0}^{2} | + |[\frac{t^2}{2} – 2 t]_{2}^{4} |$

$\displaystyle S = |[( \frac{2^2}{2} – 2 \times 2 ) – 0] | + |[( \frac{4^2}{2} – 2 \times 4 ) – (\frac{2^2}{2} – 2 \times 2)] |$

$\displaystyle S = |-2| + |2|$

S = 4 m

## A particle which is experiencing a force is given by F = 3 i – 12 j undergoes a displacement d = 4 i .

Q: A particle which is experiencing a force is given by $\vec{F} = 3 \hat{i} -12 \hat{j}$ undergoes a displacement $\vec{d} = 4 \hat{i}$ . If the particle had a kinetic energy of 3 J at the beginning of the displacement , what is its kinetic energy at the end of the displacement .

(a) 9 J

(b) 15 J

(c) 12 J

(d) 10 J

Solution : $\displaystyle W = \vec{F}.\vec{d}$

$\displaystyle W = (3 \hat{i} -12 \hat{j}).(4\hat{i})$

W = 12 J

Using Work energy Theorem ,

W = K2 – K1

12 = K2 – K1

K2 = 12 + K1

As initial kinetic energy = 3 J

K2 = 12 + 3

= 15 J

Correct option is (b)

## A force acts on a 2 kg object , so that its position is given as a function of time as x = 32 + 5 . What is the work done by this force in first 5 seconds .

Q: A force acts on a 2 kg object , so that its position is given as a function of time as x = 32 + 5 . What is the work done by this force in first 5 seconds .

(a) 850 J

(b) 900 J

(c) 950 J

(d) 875 J

Solution : x = 32 + 5

$\displaystyle \frac{dx}{dt} = 6 t$

v = 6 t

at t = 0 , v1 = 0

at t = 5 , v2 = 6 x 5 = 30 m/s

Work done = change in Kinetic Energy

$\displaystyle W = \frac{1}{2}m v_2^2 – \frac{1}{2}m v_1^2$

$\displaystyle W = \frac{1}{2}m (v_2^2 – v_1^2)$

$\displaystyle W = \frac{1}{2} \times 2 (30^2 – 0)$

W = 900 J

## A body of mass m = 10-2 kg is moving in a medium and experiences a frictional force F = -k v2 . Its initial speed is vo = 10 m/s ….

Q: A body of mass m = 10-2 kg is moving in a medium and experiences a frictional force F = -k v2 . Its initial speed is vo = 10 m/s . If after 10 sec its energy is $\frac{1}{8}m v_0^2$ , the value of k will be

(a) 10-3 kg/s

(b) 10-4 kg/m

(c) 10-1 kg/m-s

(d) 10-3 kg/m

Solution : Given F = -k v2

Acceleration $\displaystyle a = \frac{F}{m}$

$\displaystyle a = -\frac{k v^2}{m}$

$\displaystyle \frac{dv}{dt} = -\frac{k v^2}{m}$

$\displaystyle \frac{dv}{v^2} = -\frac{k}{m} dt$

$\displaystyle \int_{10}^{v} \frac{dv}{v^2} = -\frac{k}{m} \int_{0}^{t} dt$

$\displaystyle [-\frac{1}{v}]_{10}^{v} = -\frac{k}{m} t$

$\displaystyle [\frac{1}{v} – \frac{1}{10}] = \frac{k}{m} t$

$\displaystyle \frac{1}{v} = 0.1 + \frac{k}{m} t$

$\displaystyle v = \frac{1}{0.1 + \frac{k t}{m}}$

$\displaystyle v = \frac{1}{0.1 + \frac{k \times 10}{10^{-2}} }$

$\displaystyle v = \frac{1}{0.1 + 1000 k}$

$\displaystyle \frac{1}{2}m v^2 = \frac{1}{8} m v_o^2$

$\displaystyle v = \frac{v_o}{2} = \frac{10}{2}$

v = 5 m/s

$\displaystyle 5 = \frac{1}{0.1 + 1000 k}$

1 = 0.5 + 5000 k

k = 10-4 kg/m

Correct option is (b)

## A time dependent force F = 6 t acts on a particle of mass 1 kg . If the particle starts from rest , the work done by the force …

Q: A time dependent force F = 6 t acts on a particle of mass 1 kg . If the particle starts from rest , the work done by the force during the first 1 sec will be

(a) 22 J

(b) 9 J

(c) 18 J

(d) 4.5 J

Solution : Since , $\displaystyle \frac{dp}{dt} = F$

$\displaystyle dp = F dt$

$\displaystyle p = \int_{0}^{1} 6t dt$

$\displaystyle p = [3 t^2]_{0}^{1}$

p = 3 kg m/s

Kinetic Energy $\displaystyle K = \frac{p^2}{2m}$

$\displaystyle K = \frac{3^2}{2 \times 1}$

K = 4.5 J

Correct option is (d)

## The original length of spring is 25 cm . It elongates 2 cm if a force of 0.96 N is exerted on it . A container is filled with water ….

Q: The original length of spring is 25 cm . It elongates 2 cm if a force of 0.96 N is exerted on it . A container is filled with water and one end of the spring is fixed 30 cm above the surface of water in the container . A wooden block of mass 32 g and of density 0.4 g/cm3 is hanged onto the other end of the spring .The height of block is 5 cm . To what depth (in mm) does the block sink into the water ? [Take g = 9.8 m/s3/2] ## The graph shows the potential energy of an electric dipole that oscillates between ±60° . What is dipoles kinetic energy …

Q: The graph shows the potential energy of an electric dipole that oscillates between ±60° . What is dipoles kinetic energy (in μ J) when it is aligned with the field . Sol: $\displaystyle W = pE cos60^o$

$\displaystyle W = – 2 \times \frac{1}{2}$

W = – 1 Joule

– 2 + K = -1

K = 1 joule

## A metal wire PQ of mass 10 gm lies at rest on two horizontal metal rails separated by 4.90 cm . A vertically downward magnetic field …

Q: A metal wire PQ of mass 10 gm lies at rest on two horizontal metal rails separated by 4.90 cm . A vertically downward magnetic field of magnitude 0.800 T exists in the space . The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20 ohm , the wire PQ starts sliding on the rails . Find the coefficient of friction . Neglect magnetic force acting on wire PQ due to metal rails (g = 9.8 m/s2) . Sol: Force $\displaystyle \vec{F} = I [\vec{l} \times \vec{B}]$

$\displaystyle \vec{F} = I [-l \hat{j} \times -B \hat{k}]$

$\displaystyle \vec{F} = I l B \hat{i}$

Frictional force f = μ N = μ m g (Towards left)

I l B = μ m g

$\displaystyle \mu = \frac{I l B}{m g}$

$\displaystyle I = \frac{E}{R} = \frac{6}{20}$

I = 0.3 A

l = 4.9 × 10-2 m , B = 0.8 T

$\displaystyle \mu = \frac{0.3 \times 4.9 \times 10^{-2} \times 0.8}{10 \times 10^{-3} \times 9.8}$

μ = 0.12

## One end of a cylindrical rod is grounded to hemispherical surface of radius R= 20 mm . It is immersed in water (μ = 4/3) .

Q: One end of a cylindrical rod is grounded to hemispherical surface of radius R= 20 mm . It is immersed in water (μ = 4/3) . If the refractive index of the rod is 1.5 and an object is placed in water on the axis at a distance of 10 cm from the pole , then determine the position of the image from the pole on the axis (in cm)

Sol: R = 2 cm

$\displaystyle \frac{\mu_2}{v} – \frac{\mu_1}{u} = \frac{\mu_2 – \mu_1}{R}$

$\displaystyle \frac{1.5}{v} – \frac{1.33}{10} = \frac{1.5 – 1.33}{2}$

v = 3- 1.25 cm