Physics Test

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Physics Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

A point object is released at t = 0, from a point A (0, 24 m) as shown in the figure. The plane mirror is fixed at angle 45o from vertical. Distance between the object and image will be \(4 \sqrt{2} \mathrm{m}\)at
(use g = 10 ms-2)

A
t = 3s

B
\(t=\sqrt{3} s\)

C
t = 2s

D
\(t=2 \sqrt{6} s\)

Solution

Let at any time \(t,\) object is at \({ }^{\prime} \mathrm{P}^{\prime}\) and \(\mathrm{Op}=\mathrm{x}\) Then image will form at Q

\(\mathrm{PN}=\mathrm{NQ}\)
\(P N=N Q=\frac{x}{\sqrt{2}}\)
\(\mathrm{PQ}=\mathrm{PN}+\mathrm{NQ}=\sqrt{2 \mathrm{x}}\)
Now when \(P Q=4 \sqrt{2}\)
\(\Rightarrow \sqrt{2} \mathrm{x}=4 \sqrt{2}\)
\(\Rightarrow[x=4\)
Now, \(A P=24-4=20 \mathrm{m}\)
Time taken by object to come at \({ }^{\prime} \mathrm{P}^{\prime}=\mathrm{t}\) \(\mathrm{So}, \mathrm{AP}=1 / 2 \mathrm{gt}^{2}\)
\(\Rightarrow 20=1 / 2 \mathrm{gt}^{2}\)
\(\Rightarrow t=2\) second
Hence, the correct option is C.
Question 2
1 / -0

A car completes a certain journey in 8 hours. It covers half the distance at 40 km/hr and the rest at 60 km/hr. The length of the journey is...

A
350 km

B
420 km

C
384 km

D
400 km

Solution

Let the total journey be x lem.
\(\frac{x}{2} \times \frac{1}{40}+\frac{x}{2} \times \frac{1}{60}=8\)
Therefore, \(\frac{x}{8}+\frac{x}{120}=8\)
\(3 \pi+2 x=1920\)
That is, \(5 \mathrm{x}=\mathrm19 20, \mathrm{x}=384 \mathrm{km}\)
Hence, The length of the journey is \(384 \mathrm{km}\)
Hence, the correct option is C.
Question 3
1 / -0

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

A
tan⁻¹(1/3)

B
tan⁻¹(2/3)

C
tan⁻¹(3/2)

D
tan⁻¹(3/5)

Solution

\(R=6+18=24 m=\frac{\mathrm{u}^{2} \sin 2 \theta}{g} \Rightarrow u^{2}=\frac{240}{2 \sin \theta \cos \theta}=\frac{120}{\sin \theta \cos \theta}\)
\(y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta} \Rightarrow 3=6 \tan \theta-\frac{360}{2 u^{2} \cos ^{2} \theta}\)
\(\Rightarrow 3=6 \tan \theta-\frac{360}{240} \tan \theta=\frac{9}{2} \tan \theta \Rightarrow \tan \theta=\frac{2}{3}\)
Hence, the correct option is B.
Question 4
1 / -0

For a given velocity, a projectile has the same range RR for two angles of projection. If t₁ and t₂ are the time of flight in the two cases, then:

A
t₁t₂∝R

B
t₁t₂∝R2

C
t₁t₂∝1/R²

D
t₁t₂∝1/R

Solution

Let the projectiles be projected at a velocity \(u\). Let the angles of projection be \(\theta_{1}\) and \(\theta_{2}\). So, from the condition provided:
\(R=\frac{u^{2}}{g} \sin 2 \theta_{1}=\frac{u^{2}}{g} \sin 2 \theta_{2}\)
since, for both the projected angles, range is same, the angles follow the relation \(\theta_{1}+\theta_{2}=90^{0},\) or, \(\theta_{2}=90^{\circ}-\theta_{1}\)
Now, the times of flight, \(t_{1}\) and \(t_{2}\) may be expressed as:
\(t_{1}=\frac{2 u \sin \theta_{1}}{g}\) and \(t_{2}=\frac{2 u \sin \theta_{2}}{g}=\frac{2 u \cos \theta_{1}}{g}\)
So, \(t_{1} t_{2}=\frac{2 u \sin \theta_{1}}{g} \times \frac{2 u \cos \theta_{1}}{g}=\frac{2 u^{2}}{g^{2}} \times\left(2 \sin \theta_{1} \cos \theta_{1}\right)=\frac{2 u^{2}}{g^{2}} \sin 2 \theta_{1}=\frac{2 R}{g}\)
\(\Rightarrow t_{1} t_{2} \propto R\)
Hence, the correct option is B.
Question 5
1 / -0

Velocity-time graph of a particle moving in a straight line is shown in the figure. Find the total displacement of the particle.

A
40 m

B
60 m

C
80 m

D
100 m

Solution

Displacement= Area under velocity-time graph Hence, \(s_{O A}=\frac{1}{2} \times 2 \times 10=10 \mathrm{m}\)
\(s_{A B}=2 \times 10=20 \mathrm{m}\)
or \(s_{O A B}=10+20=30 \mathrm{m}\)
\(s_{B C}=\frac{1}{2} \times 2(10+20)=30 m\)
or \(s_{O A B C}=30+30=60 \mathrm{m}\)
and \(s_{O A B C D}=60+20=80 \mathrm{m}\)
Hence, the correct option is C.
Question 6
1 / -0

Wind is blowing west to east along two parallel tracks. Two trains moving with same speed in opposite directions have the relative velocity with respect to wind in the ratio 1:2. The speed of each train is

A
equal to that of wind

B
double that of wind

C
three times that of wind

D
half that of wind

Solution

Let v be velocity of wind and u be velocity of each train.
Relative velocity of one train with respect to wind =2× Relative velocity of other train with respect to wind
u+v=2(u−v)
v+2v=2u−u=u
i.e., u=3v.
Hence, the correct option is C.
Question 7
1 / -0

A body is released from the top of the tower of height H. It takes t time to reach the ground. The position of body t/2 time after release will be

A
at H/2 metres from ground

B
at H/4 metres from ground

C
at 3H/4 metres from the ground

D
at H/6 metres from the ground

Solution

\(H=\frac{1}{2} g t^{2}\)
At \(\frac{t}{2}, s=\frac{1}{2} g(t / 2)^{2}=\frac{1}{4}\left(\frac{1}{2} g t^{2}\right)\)
\(s=\frac{1}{4} H\)
Therefore, the height from ground is given by, \(x=H-s=H-\frac{1}{4} H=\frac{3}{4} H\)
Hence, the correct option is C.
Question 8
1 / -0

A particle aimed at a target, projected with an angle 15o with the horizontal is short of the target by 10 m . If projected with an angle of 45o is away from the target by 15 m, then the traget by 15 m, then the angle of projection to hit the target is:

A
1/2 sin⁻¹ (1/10)

B
1/2 sin⁻¹(3/10)

C
12 sin⁻¹(9/10)

D
1/2 sin⁻¹(7/10)

Solution

\(R_{1}=R-10\)
\(R_{2}=R+15\)
\(\frac{R-10}{R+15}=\frac{\sin 2 \theta_{1}}{\sin 2 \theta_{2}}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}=\frac{1}{2}\)
\(2 R-20=R+15\)
\(\Rightarrow R=35\)
\(R_{\max }=\frac{u^{2}}{g} \Rightarrow 50=\frac{u^{2}}{g} \Rightarrow u^{2}=50 g\)
\(R=\frac{u^{2} \sin 2 \theta}{g}\)
\(35=\frac{50 g \times \sin 2 \theta}{g}\)
\(\sin 2 \theta=\frac{35}{50}=\frac{7}{10}\)
\(2 \theta=\sin ^{-1}\left[\frac{7}{10}\right]\)
\(\theta=\frac{1}{2} \sin ^{-1}\left[\frac{7}{10}\right]\)
Hence, the correct option is D.
Question 9
1 / -0

A police party is moving in a jeep at a constant speed V. They saw a thief at a distance x on a motorcycle which is at rest. The moment the police saw the thief, the thief started at constant acceleration a. Which of the following relations is true if the police is able to catch the thief?

A
V²

B
V²<2ax

C
V²≥2ax

D
V²=ax

Solution

Let the time elasped for catching the thief is \(t\) and distance travelled by thief is \(y\). \(\Rightarrow y=\frac{1}{2} a t^{2}\)
\(\Rightarrow x+y=V t\)
Substituting \(t=\frac{x+y}{V}\) in the first equation we get
\(\Rightarrow a(x+y)^{2}=2 V^{2} y\)
By \(A M-G M\) for LHS we get \(\Rightarrow(x+y)^{2} \geq 2 x y\)
\(\Rightarrow \frac{2 V^{2} y}{a} \geq 2 x y\)
\(\Rightarrow V^{2} \geq 2 a x\)
Hence, the correct option is C.
Question 10
1 / -0

A bird files with a speed v=|t−2|m/s along a straight line, where t is the time in seconds. The distance travelled by the bird during first four seconds is equal to

A
2 m

B
4 m

C
6 m

D
8 m

Solution

For \(t=0\) to \(t=2\)
\(v=-t+2\)
For \(t=2\) to \(t=4\)
\(v=t-2\)
Area of the graph of \(v v / s t\) is \(\left(\frac{1}{2} \times 2 \times 2\right) \times 2=4 m\)
Hence, the correct option is B.

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Physics Test - 17

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Physics Test - 17

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Question 1

t = 3s

\(t=\sqrt{3} s\)

t = 2s

\(t=2 \sqrt{6} s\)

Solution

Question 2

350 km

420 km

384 km

400 km

Solution

Question 3

tan−1(1/3)

tan−1(2/3)

tan−1(3/2)

tan−1(3/5)

Solution

Question 4

t1t2∝R

t1t2∝R2

t1t2∝1/R2

t1t2∝1/R

Solution

Question 5

40 m

60 m

80 m

100 m

Solution

Question 6

equal to that of wind

double that of wind

three times that of wind

half that of wind

Solution

Question 7

at H/2 metres from ground

at H/4 metres from ground

at 3H/4 metres from the ground

at H/6 metres from the ground

Solution

Question 8

1/2 sin−1 (1/10)

1/2 sin−1(3/10)

12 sin−1(9/10)

1/2 sin−1(7/10)

Solution

Question 9

V2

V2<2ax

V2≥2ax

V2=ax

Solution

Question 10

2 m

4 m

6 m

8 m

Solution

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tan⁻¹(1/3)

tan⁻¹(2/3)

tan⁻¹(3/2)

tan⁻¹(3/5)

t₁t₂∝R

t₁t₂∝R2

t₁t₂∝1/R²

t₁t₂∝1/R

1/2 sin⁻¹ (1/10)

1/2 sin⁻¹(3/10)

12 sin⁻¹(9/10)

1/2 sin⁻¹(7/10)

V²

V²<2ax

V²≥2ax

V²=ax