Motion in A Straight Line Test

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Motion in A Straight Line Test - 55

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Question 1
1 / -0

If a car at rest accelerates uniformly to a speed of 144 km/h in 20s , it covers a distance of

A
400 m

B
1440 m

C
2880 m

D
none of these

Solution
Question 2
1 / -0

A ball is thrown upwards from the ground with an initial velocity u. The ball is at a height of $$80$$m at two times. The time interval being $$6s$$. Take $$g=10m/s^2$$, the value of u is

A
$$100$$m/s

B
$$80$$m/s

C
$$50$$m/s

D
$$30$$m/s

Solution

From situation $$B$$ to $$C$$
$$t=6\ s$$
$$\implies \cfrac{2v}{g}=6$$
$$\implies v=\cfrac{6\times 10}{2}=30\ m/s$$
Now from $$A$$ to $$B$$
$$v^2=u^2+2as$$
$$\implies 30^2=u^2+2(-10)(80)$$
$$\implies u=50\ m/s$$
Question 3
1 / -0

A laser gun of power $$3 \,mW$$ and mass $$50 \,gm$$ emits photons of wavelength $$500 \,nm$$. It is gravity frequency space and emits for one hour. Find the distance covered by gun due to recoil in this one hour (approx).

A
$$1.3 \,mm$$

B
$$2.1 \,\mu m$$

C
$$1.1 \,cm$$

D
$$8.5 \,mm$$

Solution

$$S = \dfrac{1}{2}at^2$$ $$t = 1 \,hr = 3600 \,sec$$
Let us find force on the gun
$$=$$ no. of photon emitted $$\times$$ change in momentum

$$= \left(\dfrac{Total \,energy \,emitted}{Energy \,of \,photon}\right) \times \left(\dfrac{h}{\lambda}\right)$$

$$= \dfrac{3 \times 10^{-3} Joules}{\dfrac{hc}{\lambda}} \times \dfrac{h}{\lambda}$$

$$= \dfrac{3 \times 10^{-3}}{3 \times 10^8} = 10^{-11} N$$

$$\therefore a = \dfrac{Force}{Mass} = \dfrac{10^{11}}{\left(\dfrac{50}{1000}\right)} = 20 \times 10^{-11} \,m/s^2$$

$$\therefore S = \dfrac{1}{2} \times 20 \times 10^{-11} \times (3600)^2$$

$$= 10^{-10} \times 12.96 \times 10 \,m$$
$$= 1.296 \,mm$$
$$= 1.3 \,mm$$
Question 4
1 / -0

A block of mass $$5Kg$$ is kept on the floor of an elevator kept at rest. The elevator started going down with an acceleration of $$14{ ms }^{ -2 }$$ at time $$t=0$$. The magnitude of displacement of the block during $$t=0$$ to $$t=0.1s$$ is $$\left[ g=10{ ms }^{ -2 } \right]$$

A
$$\dfrac { 1 }{ 20 } m$$

B
$$20 /m$$

C
$$\dfrac { 7 }{ 20 } m$$

D
$$\dfrac { 12 }{ 20 } m$$

Solution
Question 5
1 / -0

A train is moving along a straight path with uniform acceleration its engine passes across a pole with a velocity of $$60km/h$$ and the end (guard'svan) passes across same pole with a velocity of $$80km/h$$. The middle point of the train will pass across same pole with a velocity

A
$$ 90km/h$$

B
$$70.7 m/h$$

C
$$65km/h$$

D
$$75km/h$$

Solution

Let S be the length of the train t be the time taken by the full train to cross the pole and a be uniform acceleration of train from -
$$\Rightarrow v^2-u^2=2as$$
$$\Rightarrow \dfrac{80^2-60^2}{2a}=s$$

$$\Rightarrow s=\dfrac{6400-3600}{2a}$$
$$=\dfrac{1400}{2a}\longrightarrow \left( 1\right)$$
The middle part of the train covers half of total distance $$\dfrac{s}{2}=\dfrac{700}{a}$$ from $$v^2-u^2=2as,$$
$$\Rightarrow v^2-60^2=2a\times \dfrac{70}{a}=1400$$
$$v^2=1400+3600=\sqrt{5000}=70.7km/hr$$
Hence, the answer is $$70.7km/hr.$$
Question 6
1 / -0

Find the odd one out and give the reason : speed, distance , mass, velocity

A
Mass

B
Speed

C
Distance

D
Velocity

Solution
Question 7
1 / -0

A body travelling with uniform acceleration crosses two points $$A$$ and $$B$$ with velocities $$20\ m/sec$$ and $$30\ m/sec$$ respectively then the speed of the body at mid point of $$A$$ and $$B$$ is

A
$$25\ m/sec$$

B
$$25.5\ m/sec$$

C
$$24\ m/sec$$

D
$$10\sqrt { 6 }\ m/sec$$

Solution

Let the acceleration of the car is $$= a$$ and the distance between $$A $$ and $$B$$ is $$=d$$

$$v^2-u^2=2ad$$

Given,

$$v=30\,m/s$$ and $$u=20\,m/s$$

$$2ad=30^2-20^2$$

$$ad=\dfrac{(900-400)}{2}=250$$

when the car is at the mid point of $$AB$$ then let the speed of the car is
$$v_1$$

$$v_1^2-20^2=2a(\dfrac d2)$$

$$v_1^2=ad+400=250+400=650$$

therefore

$$v_1$$ is $$= 25.5\,m/s$$
Question 8
1 / -0

A ball dropped from some height coves half of its total during the last second of its free fall. Find

A
time of flight

B
height of its fall

C
speed with which it strikes the ground.

D
none of these
Question 9
1 / -0

A particle falling from rest under gravity covers a distance $$x$$ in $$4\ s$$. If it continues falling then next $$2\ x$$ distance will be covered in approximately

A
$$1.41\ s$$

B
$$1.73\ s$$

C
$$2.05\ s$$

D
$$2.92\ s$$

Solution

$$ h = \frac{1}{2} gt^{2}$$
given $$x = \frac{1}{2} g (4)^{2} = 89 $$ ....(1)
Next 2x means total x+2x=3x from rest
so just use $$ h=\frac{1}{2}gt^{2}$$
& put h = 3x
we get $$ 3x = \frac{gt^{2}}{2}$$
$$ \Rightarrow t^{2} = \frac{6x}{9}$$
$$ \Rightarrow t = \sqrt{\frac{6x}{9}} = \sqrt{\frac{6}{9} \times 89}$$ ; from equation
$$ \Rightarrow t = \sqrt{48} = 4\sqrt{3} = 4 \times 1.732 = 6.928 sec$$
So time for distance b/w x & 3x
will be 6.928 - 4 = 2.928 sec
Question 10
1 / -0

A body released from the top of alls through a height of $$5 \mathrm { m }$$ during the first second of its fall and $$35 \mathrm { m }$$ during the last second of its fall. The height of the tower is

A
$$80 \mathrm { m }$$

B
$$60 \mathrm { m }$$

C
$$40 \mathrm { m }$$

D
$$20 \mathrm { m }$$

Solution