Motion in A Straight Line Test

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Motion in A Straight Line Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

A free falling body travels ___ of total distance in 5th second.

A
8 %

B
12 %

C
25 %

D
36 %

Solution

The correct option is D

Given,

A free falling body

Distance traveled in 5s

$$S=ut+\dfrac{1}{2}gt^2$$ since $$u=0,g=10m/s$$

$$=\dfrac{1}{2}\times10\times5^2$$

$$=125m$$

Distance travelled in 5^th sis:

Distance travelled in 5s- Distance travelled in 4s

Distance travelled in $$4s=\dfrac{1}{2}\times10\times4^2$$

$$=80m$$

So,

$$s_5-s_4=125-80=45$$

Therefore,

$$\dfrac{s_5-s_4}{s_5}\times100=\dfrac{45}{125}\times100=36\%$$
Question 2
1 / -0

A body of mass $$m$$ accelerates uniformly from rest to velocity $${v}_{1}$$ in time interval $${T}_{1}$$. The instantaneous power delivered to the body as a function of time $$t$$ is:

A
$$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } t$$

B
$$\dfrac { { mv }_{ 1 } }{ { T }_{ 1 }^{ 2 } } t$$

C
$${ \left( \dfrac { { mv }_{ 1 } }{ { T }_{ 1 } } \right) }^{ 2 }t$$

D
$$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 } } { t }^{ 2 }$$

Solution

Let a be the constant acceleration
Using 1st equation of motion
$$v_1=u+aT_1$$
$$v_1=0+aT_1$$
$$a=\dfrac{v_1}{T_1}..........(i)$$
At any instant t, the velocity v of the body
$$v=0+at$$
$$v=\dfrac{v_1}{T_1}\cdot t.........(ii)$$
$$\because P=Fv$$
$$P=mav$$
$$P=m\left(\dfrac{v_1}{T_1}\right)\left(\dfrac{v_1}{T_1}\cdot t\right)$$
$$P=\dfrac{mv_1^2}{T_1^2}t$$
Question 3
1 / -0

Directions For Questions

A car accelerates from rest at constant rate of $$2\ {ms}^{-2}$$ for some time. Immediately after this, it retards at a constant rate of $$4\ {ms}^{-2}$$ and comes to rest. The total time for which it remains in motion is $$3\ s$$. Taking the moment of start of motion as $$t=0,$$ answer the following question.

...view full instructions

If the time of acceleration is $${t}_{1}$$, then the speed of the car at $$t={t}_{1}$$ is:

A
$$2\ {t}_{1}$$

B
$$4\ {t}_{1}$$

C
$$>2\ {t}_{1}$$

D
$$<2\ {t}_{1}$$

Solution

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Question 4
1 / -0

A balloon is ascending vertically with an acceleration of 0.2 m $$s^{-2}$$ . Two stones are dropper from it at an interval of 2 s. The distance between them when the second stone is dropped is (take g = 9.8 m $$s^{-2}$$):

A
24 m

B
4.9 m

C
19.6 m

D
20.0 m

Solution

A balloon is ascending vertically with an acceleration= $$0.2 m/s^2$$
Two stones are dropped from it at an interval of $$t=2s$$ .
$$g=9.5 m/S$$
The distance between them when the $$2nd$$ stone is dropped:
$$s_1=\cfrac {1}{2}at^2$$ $$(\because u=0)$$
or, $$s_1=\cfrac {1}{2}\times 0.2\times 2^2$$
$$=\cfrac {1}{2}\times \cfrac {2}{10}\times 2^2$$
$$=\cfrac {4}{10}=0.4 m$$ .
Due to free fall the $$1st$$ stone will have,
$$s=\cfrac {1}{2}\times (0.2+9.8)\times 2^2$$
$$=\cfrac {1}{2}\times 10 \times 4$$
$$=20$$ .
$$\therefore$$ Hence the $$2nd$$ stone is dropped= $$(20+4)=24 m$$.
Question 5
1 / -0

The velocity acquired by a body moving with uniform acceleration is $$30 \,m/s$$ in $$2$$ seconds and $$60 \,m/s$$ in $$4$$ seconds. The initial velocity is:

A
$$Zero$$

B
$$2 m/s$$

C
$$4 m/s$$

D
$$10 m/s$$

Solution

Let,

Constant acceleration, $$a$$

Initial velocity $$u$$

Apply kinematic Equation

$$v=u+at$$

$$a=\dfrac{v-u}{t}$$

Compare acceleration at both event at time, $${{t}_{1}}\,\And \,{{t}_{2}}$$

$$\dfrac{{{v}_{2}}-u}{{{t}_{1}}}=\dfrac{{{v}_{3}}-u}{{{t}_{2}}}$$

$$\Rightarrow \dfrac{30-u}{2}=\dfrac{60-u}{4}$$

$$\Rightarrow u=0$$

Initial velocity $$u=zero$$
Question 6
1 / -0

A body freely falling from a height $$h$$ describes $$\dfrac {7h}{16}$$ in the last second of its fall. The height $$h$$ is $$(g=10\ {ms}^{-2})$$

A
$$40\ m$$

B
$$45\ m$$

C
$$80\ m$$

D
$$160\ m$$

Solution

$$h=ut+\dfrac{1}{2}gt^2$$
u=0
$$h=\dfrac{1}{2}gt^2$$
distance covered in $$t^{th}$$ sec =$$s_t=\dfrac{7g}{16}=u+\dfrac{a}{2}(2t-1)=\dfrac{a}{2}(2t-1)$$
solving both the equations we get
$$7t^2-32t+16=0$$
$$t=4secs$$
h=$$\dfrac{1}{2}g4^2=80m$$
Question 7
1 / -0

A satellite orbiting round the earth appears stationary when:

A
Its time period is one day and it is rotating in the same sense as that of earth.

B
Its time period is one day and it is rotating normal to the direction of earth.

C
Its time period is $$12\ hr$$ and it is rotating in the same direction as that of earth.

D
Its time period is $$12\ hr$$ and it is rotating normal to earth.

Solution

a satellite orbit revolving around the earth has same time period as that of the earth that is 1 day and it rotates in the same direction as that of the earth
so the relative velocity of the satellite with respect to earth is zero
so it appears to be stationary to earth
Question 8
1 / -0

If a particle is thrown with velocity more than $$10\ m/s$$ vertically upward, then the distance traveled by the particle in last second of its ascent is:

A
$$g$$

B
$$\dfrac { g }{ 2 }$$

C
$$\dfrac { g }{ 4 }$$

D
$$\dfrac { g }{ 8 }$$

Solution

time of ascent be t
$$0=10-gt$$
$$t=1s$$
$$s=ut-\dfrac{1}{2}gt^2=\dfrac{g}{2}$$
Question 9
1 / -0

A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the shortest distance of the block from the point where the force started acting.

A
10 m

B
8 m

C
6 m

D
2 m

Solution

Assume initial velocity of 1.5 m/s is in the x-direction
Since there are no forces on it in this direction, there will be no acceleration.
So, distance $$S_x=1.5\times 4=6m$$
In the y-direction, $$F=5N$$ and $$m=5kg$$
Acceleration in y=direction,
$$a_y=\dfrac{F}{m}=\dfrac{5}{5}=1 m/s^2$$
$$S_y=\dfrac{1}{2}a_yt^2=\dfrac{1}{2}\times 1\times 4^2=8m$$
Resolving the x and y vector we get,
$$S^2=S_x^2+S_y^2$$
$$S^2=6^2+8^2$$
$$S=\sqrt{36+64}$$
$$S=\sqrt{100}$$
$$S=10m$$
Question 10
1 / -0

The speed of a body is doubled when it moves over a distance of $$10\ m$$. If the initial speed be $$u$$, what will be the speed after further coverage of distance $$10\ m$$?

A
$$\sqrt 5 u$$

B
$$\sqrt 6 u$$

C
$$\sqrt 7 u$$

D
$$\sqrt 8 u$$

Solution

If the acceleration be $$a $$ then using the third equation of motion, $$v^2=u^2+2as$$
we have $$(2u)^2=u^2 +2a\times 10$$ so $$a=3u^2/20$$....(1)
After further covering 10 m taking initial speed at the $$moment $$ it has covered the $$first $$ $$10m$$ as $$2u$$ we get $$v^2=(2u)^2 +2a\times 10$$ , putting value of $$a $$ from equation-1 we get $$v^2=\sqrt{7}u$$
Option C is correct.

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 48

Result

Motion in A Straight Line Test - 48

Score

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Rank

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SHARING IS CARING

Question 1

8 %

12 %

25 %

36 %

Solution

Question 2

$$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } t$$

$$\dfrac { { mv }_{ 1 } }{ { T }_{ 1 }^{ 2 } } t$$

$${ \left( \dfrac { { mv }_{ 1 } }{ { T }_{ 1 } } \right) }^{ 2 }t$$

$$\dfrac { { mv }_{ 1 }^{ 2 } }{ { T }_{ 1 } } { t }^{ 2 }$$

Solution

Question 3

$$2\ {t}_{1}$$

$$4\ {t}_{1}$$

$$>2\ {t}_{1}$$

$$<2\ {t}_{1}$$

Solution

Question 4

24 m

4.9 m

19.6 m

20.0 m

Solution

Question 5

$$Zero$$

$$2 m/s$$

$$4 m/s$$

$$10 m/s$$

Solution

Question 6

$$40\ m$$

$$45\ m$$

$$80\ m$$

$$160\ m$$

Solution

Question 7

Its time period is one day and it is rotating in the same sense as that of earth.

Its time period is one day and it is rotating normal to the direction of earth.

Its time period is $$12\ hr$$ and it is rotating in the same direction as that of earth.

Its time period is $$12\ hr$$ and it is rotating normal to earth.

Solution

Question 8

$$g$$

$$\dfrac { g }{ 2 }$$

$$\dfrac { g }{ 4 }$$

$$\dfrac { g }{ 8 }$$

Solution

Question 9

10 m

8 m

6 m

2 m

Solution

Question 10

$$\sqrt 5 u$$

$$\sqrt 6 u$$

$$\sqrt 7 u$$

$$\sqrt 8 u$$

Solution

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