MCQ

MCQ 11 Mark

At the maximum height, a body thrown vertically upwards has:

A
velocity not zero but acceleration zero.
✓
acceleration not zero but velocity zero.
C
both acceleration and velocity are zero.
D
both acceleration and velocity are not zero.

Answer

Correct option: B.

acceleration not zero but velocity zero.

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MCQ 21 Mark

The distance covered in adjoining velocity - time graph is :

A
25 m
B
40 m
✓
50 m
D
45 m

Answer

Correct option: C.

50 m

(c)
50 m
Explanation:
Distance covered = Area under velocity - time graph
$
\begin{array}{l}
=1 / 2 \times \text { base } \times \text { height } \\
a=1 / 2 \times 5 \times 20=50 m
\end{array}
$

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MCQ 31 Mark

In velocity time graph, the acceleration is

✓
$-4 m / s ^2$
B
$4 m / s ^2$
C
$10 m / s ^2$
D
zero

Answer

Correct option: A.

$-4 m / s ^2$

(a)
$-4 m / s ^2$
Explanation :
Acceleration = Slope of velocity time graph
$
\begin{array}{l}
a=-\frac{20}{5} \quad[\because \text { Velocity is decreasing with time }] \\
a=-4 m / s^2[\because \text { Acceleration will be negative }]
\end{array}
$

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MCQ 41 Mark

A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from $54 km / h$ to $18 km / h$ in 5 s . What is the distance travelled by the train during this interval of time?

A
52 m
✓
50 m
C
25 m
D
80 m

Answer

Correct option: B.

50 m

(b)
50 m
Explanation :
$
\begin{array}{l}
u=54 km / h=54 \times \frac{5}{18} m / s=15 m / s \\
v=18 km / h=18 \times \frac{5}{18} m / s=5 m / s \\
t=5 S ; S=\text { ? } \\
a=\frac{v-u}{t}=\frac{5-15}{5}=-2 m / s^2 \\
v^2-u^2=2 a S \\
(5)^2-(15)^2=2 a S \\
S=\frac{25-225}{-4}=\frac{-200}{-4}=50 m
\end{array}
$

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MCQ 51 Mark

What does the area of an acceleration - time graph represent?

✓
Uniform velocity
B
Displacement
C
Distance
D
Variable velocity

Answer

Correct option: A.

Uniform velocity

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MCQ 61 Mark

A body projected vertically up with a velocity $10 m / s$ reaches a height of 20 m . If it is projected with a velocity of $20 m / s$, then the maximum height reached by the body is :

A
20 m
✓
10 m
C
80 m
D
40 m

Answer

Correct option: B.

10 m

(b) 10 m
Explanation:
Initial velocity $=u=10 m / s$
Height $=$ Distance $= S =20 m$
At highest point, Final velocity $=v=0$
Acceleration $=(a)=$ ?
$
v^2-u^2=2 a S ;(0)^2-(10)^2=2 a(20)
$
$
a=-\frac{100}{40}=-2.5 ms^{-2}
$
Case-II
$
\begin{array}{l}
u=20 m / s ; v=0 \\
a=-2.5 ms^{-2} ; S=h=? v^2-u^2=2 a S \\
(0)^2-(20)^2=2(-2.5) h \\
h=\frac{400}{5}=80 m
\end{array}
$

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MCQ 71 Mark

The speed of a car reduces from $15 m / s$ to $5 m / s$ over a displacement of 10 m . The uniform acceleration of the car is :

✓
$-10 m / s ^2$
B
$+10 m / s ^2$
C
$2 m / s ^2$
D
$0.5 m / s ^2$

Answer

Correct option: A.

$-10 m / s ^2$

(a) $-10 m / s ^2$
$\begin{array}{l}\text { Explanation: } \\ \text { Initial velocity }= u =15 m / s \\ \text { Final velocity }= v =5 m / s \\ \text { Displacement }= s =10 m \\ \text { Acceleration }= a =? \\ v ^2- u ^2=2 aS ;(5)^2-(15)^2=2 a (10) \\ 25-225=20 a \\ a =-220 / 20=-10 ms^{-2}\end{array}$

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MCQ 81 Mark

A body dropped from the top of a tower reaches the ground in 4 s . Height of the tower is