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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 2
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Q.1
At a metro station, a girl walks up a stationary escalator in time a. If she remains stationary on the escalator, then the escalator take her up in time b. The time taken by her to walk up on the moving escalator will be
0%
$\frac {a+b}{2}$
0%
a+b
0%
$\frac {ab}{a- b}$
0%
$\frac {ab}{a+ b}$
Explanation
Solution:
Let the escalator's length be D. Walking up a stationary escalator in time a means her walking speed (relative to the steps) is D/a. Standing still while the escalator carries her up in time b means the escalator's own speed is D/b.
Walking on the moving escalator, the two speeds add: D/a + D/b = D(a+b)/(ab). Time taken = D ÷ [D(a+b)/(ab)] = ab/(a+b).
Q.2
Which of the following option is correct for the object having a straight line motion represented by the following graph
0%
average velocity is zero
0%
velocity of the object increases uniformly
0%
the object moves with constantly increasing velocity from O to A and then it moves with constant velocity
0%
the graph shown is impossible
Explanation
Solution:
This is a time-vs-displacement graph (time on the vertical axis, displacement s on the horizontal axis), and the curve leaves the origin O, bulges out to positive s, then curves back to point D directly above O -- meaning D has the same s-coordinate (s=0) as the start.
Since the object's net displacement from start to end is zero (it returns to s=0), its average velocity -- displacement over the total time -- is zero, regardless of the path it followed getting there and back.
Q.3
A body starts from rest at time t=0, the acceleration- time graph is shown in below figure. The maximum velocity attained by the body will be
0%
110 m/s
0%
550 m/s
0%
650 m/s
0%
55 m/s
Explanation
Solution:
The maximum velocity is reached exactly when the acceleration drops to zero (after that point, deceleration would start reducing the velocity again), and velocity gained is the area under the acceleration-time graph up to that point.
The graph is a straight line from (0s, 10 m/s²) down to (11s, 0 m/s²), a right triangle. Area = ½ × base × height = ½ × 11 × 10 = 55, so the maximum velocity is 55 m/s.
Q.4
A particle moving in a straight line covers half the distance with a speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during the entire motion is
0%
4 m/s
0%
5 m/s
0%
5.5 m/s
0%
4.8 m/s
Explanation
Solution:
Let the total distance be 2d, so each half is d.
First half: distance d at 3 m/s takes d/3 s.
Second half: covered in two EQUAL time intervals (each of duration τ) at 4.5 m/s and 7.5 m/s, covering 4.5τ+7.5τ=12τ = d, so τ=d/12, and the total time for this half is 2τ=d/6.
Total time = d/3 + d/6 = d/2. Average speed = total distance ÷ total time = 2d ÷ (d/2) = 4 m/s.
Q.5
A particle moving along x-axis has acceleration f at time t, given by $f=f_0 \frac {t}{T}$,where f_0 and T are constants. The particle at t=0 has zero velocity. In the time interval between t=0 and the instant when f=0, the particle’s velocity $v_x$ is
0%
$\frac {1}{2} f_0 T^2$
0%
$f_0 T^2$
0%
$\frac {1}{2} f_0 T$
0%
$f_0 T $
Explanation
Solution:
With acceleration f=f
0
(1-t/T) (falling linearly from f
0
at t=0 to zero at t=T), the velocity gained is the area under the acceleration-time graph, a triangle of base T and height f
0
:
v = ½ × T × f
0
= ½f
0
T.
Q.6
The relation between time t and distance x is $t=ax^2+bx$,where a and b are constants. The acceleration is
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$-2abv^2$
0%
$2bv^3$
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$-2av^3 $
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$2av^2$
Explanation
Solution:
Differentiating t=ax²+bx with respect to x: dt/dx = 2ax+b, so v = dx/dt = 1/(2ax+b).
Differentiating v with respect to x: dv/dx = -2a/(2ax+b)² = -2av² (since 1/(2ax+b) = v).
Acceleration = dv/dt = (dv/dx)(dx/dt) = (-2av²)(v) = -2av³.
Q.7
A ball is dropped from a bridge 122.5 m above a river.After the ball has been falling for 2 seconds, a second ball is thrown straight down after it. What must the initial velocity of the second ball be so that both hit the water at the same time?
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40 m/s
0%
26.1 m/s
0%
9.6 m/s
0%
55.5 m/s
Explanation
Solution:
Time for the first ball to fall 122.5 m from rest: 122.5 = ½(9.8)t² ⇒ t²=25 ⇒ t=5 s.
The second ball is thrown 2 s later, so it only has 5-2=3 s to cover the same 122.5 m: 122.5 = 3v
0
+ ½(9.8)(3²) = 3v
0
+ 44.1.
3v
0
= 78.4 ⇒ v
0
≈ 26.1 m/s.
Q.8
A car is moving at a speed of 50 km/hr can be stopped by brakes after atleast 6 m.If the same car is moving with the speed of 100km/hr,the minimum stopping distance is
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12 m
0%
18 m
0%
6 m
0%
24 m
Explanation
Solution:
For constant braking deceleration, v²=2as, so the stopping distance s is proportional to the square of the speed.
Doubling the speed (50 to 100 km/hr) means the stopping distance scales by 2²=4: 6 m × 4 = 24 m.
Q.9
Two boys are standing at the ends A and B of a ground, where AB=a. The boy at B starts running in a direction perpendicular to AB with velocity The boy at A starts running simultaneously with velocity v and catches the other by in a time t, where t is
0%
$\frac {a}{\sqrt {v^2+v_1^2 }}$
0%
$\sqrt { \frac {a^2}{v^2 -v_1^2}}$
0%
$\frac {a}{v-v_1}$
0%
$\frac {a}{v+v_1}$
Explanation
Solution:
Set up coordinates with A at the origin and B at (a, 0). B runs at speed v
1
perpendicular to AB, so at time t, B is at (a, v
1
t). A runs in a single fixed direction at speed v, aiming to intercept B, so at time t, A is at (vt cosθ, vt sinθ) for some fixed angle θ.
For them to meet: vt cosθ=a and vt sinθ=v
1
t, so sinθ=v
1
/v and cosθ=a/(vt). Using sin²θ+cos²θ=1:
(v
1
/v)² + (a/(vt))² = 1 ⇒ a² = t²(v²-v
1
²) ⇒ t = √(a²/(v²-v
1
²)).
Q.10
A bus starts from rest with an acceleration of A man, who is 48 m behind the bus, starts with a uniform velocity of 10 m/s. Then, the minimum time after which the man will catch the bus is
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4 sec
0%
8 sec
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16 sec
0%
10 sec
Explanation
Solution:
With the bus starting from rest with acceleration 1 m/s² and the man 48 m behind running at a constant 10 m/s, set the man's starting point as the origin: the bus starts at x=48.
Bus position: x=48+½(1)t². Man's position: x=10t. Catching up means 10t = 48+0.5t², i.e. 0.5t²-10t+48=0, i.e. t²-20t+96=0.
Solving: t=(20±√(400-384))/2=(20±4)/2, giving t=8 or t=12 -- the man first draws level with the bus at the smaller value, t=8 s (after that the accelerating bus pulls away again until it's caught a second, later time).
Q.11
The acceleration ‘a’ in m/s
2
of a particle is given by $a=3t^2+2t+2$ where t is the time in seconds. If the particle starts out with a velocity v=2 m/s at t=0, then the velocity at the end of 2 seconds is
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12 m/s
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36 m/s
0%
27 m/s
0%
18 m/s
Explanation
Solution:
Integrating a(t)=3t²+2t+2 with respect to t gives v(t)=t³+t²+2t+C. Using v=2 at t=0 gives C=2, so v(t)=t³+t²+2t+2.
At t=2: v = 8+4+4+2 = 18 m/s.
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