Basic Science Test 11

Result

Basic Science Test 11

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Figure shows the graph of x-coordinate of a particle moving along x-axis as function of time. Average velocity during $t=0$ to $6s$ and instantaneous velocity at $t=3s$ respectively, will be

A
$10m/s,0$

B
$60m/s,0$

C
$0,0$

D
$0,10m/s$

Solution

Velocity= slope of x-t graph=0 (at $t=35$ )
Net displacement from $t=0$ to $t=6s=0$
Ave. velocity $=\cfrac{Net\quad displacement}{time}=\cfrac{0}{6}=0$
Question 2
1 / -0

A bob of a simple pendulum of length $2m$ is moved sideways and whirled in horizontal circle of radius $1.732m$ . If the mass of the bob is $2kg$ , then tension in the string is

A
$34.5N$

B
$22.8N$

C
$39.2N$

D
$57.6N$

Solution

$\sin\theta=\frac{1.732}{2}$
$\sin\theta=\frac{\sqrt{3}}{2}$
$\theta=60^0$
$T\cos\theta=mg$
$T=\frac{mg}{\cos\theta}=\frac{2*9.8}{\cos60^0}=\frac{19.6}{\frac{1}{2}}=39.8N$
Question 3
1 / -0

A particle is in uniform circular motion, then its velocity is perpendicular to

A
Net force

B
Centripetal acceleration

C
Angular velocity

D
All of these

Solution

$V\bot a_c$
if in uniform circular motion then only centripetal force occurs
net force =centripetal force
so netforce $\bot a_v$
Question 4
1 / -0

A wheel of mass $40kg$ and radius of gyration $0.5m$ comes to rest from a speed of $1800r\pm$ in $30s$ . Assuming that the retardation is uniform, the value of the retarding torque in $Nm$ , is

A
$10\pi$

B
$20\pi$

C
$30\pi$

D
$40\pi$

Solution

$\tau =I\alpha$
$=(m{k}^{2}).\alpha$
$=40\times {(0.5)}^{2}\times 2\pi$
$=20\pi$ rad/s
Question 5
1 / -0

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination $\theta$ ?

A
$\frac{2}{7}\tan\theta$

B
$\frac{1}{3}g\tan\theta$

C
$\frac{1}{2}\tan\theta$

D
$\frac{2}{5}\tan\theta$

Solution

Here we are talking about minimum coefficient of friction this suggests that friction acting would be $\mu mg\cos\theta$
taking moment about centre,
$\mu mg\cos\theta*r=I*\alpha$
or, $=\frac{\mu mg\cos\theta_{r}^2}{I}$
now, from force balance,
$mg\sin\theta-\mu mg\cos\theta=ma$
$a=g\sin\theta-\mu g\cos\theta$
from condition of rolling we know that
$a=r\alpha$
$\frac{\mu mg\cos\theta_{r}^2}{I}=g\sin\theta-\mu mg\cos\theta$
simplifying this, we get
$\mu=\frac{\tan\theta}{1+\frac{mr^2}{I}}$
$I=\frac{2}{5}mr^2$
Putting the value of moment of ionertia in expression of coefficient of friction we get,
$\mu=\frac{2}{7}\tan\theta$
Question 6
1 / -0

A body is started from rest with acceleration $2 m/s^{2}$ till it attains the maximum velocity then retards to rest with 3 $m/s^{2}$ It total time taken is 10 second then maximum speed attained is

A
$12 m/s$

B
$8 m/s$

C
$6 m/s$

D
$4 m/s$
Question 7
1 / -0

A partice is at rest and it is at origin $(0,0)$ at $t=0$ . If velocity varies as $v=k{x}^{2}$ where $v$ is the velocity and $x$ is displacement. Find how much distance it will travel in $2$ sec

A
$5m$

B
$2m$

C
$0$

D
none of these

Solution

$V=k{x}^{2}$
$\cfrac{dx}{dt}=k{x}^{2}$
$dx=k{x}^{2}dt$
$\int _{ 0 }^{ x }{ \cfrac { 1 }{ { x }^{ 2 } } } =\int _{ 0 }^{ 2 }{ x } dt$
$-\cfrac{1}{x}={ \left[ kt \right] }_{ 0 }^{ 2 }$
$-\cfrac{1}{x}=k2$
$x=-\cfrac{1}{2k}$
$-\cfrac{1}{x}=kt+k$
at $t=0$ and $d=0$
$k=\infty$
$x=-\cfrac{1}{2\infty}$
$k=0$
Question 8
1 / -0

Two identical rings each of mass m with their planes mutually perpendicular , radius R are welded at their point of contact O . If the system is free to rotate about an axis passing through the point p perpendicular to the plane of the paper the moment of inertia of the system about this axis equal to :

A
$6.5 m R^2$

B
$12m R^2$

C

$6 m R^2$

D
$11.5 m R^2$

Solution

Answer
$I = I_{1} + I_{1}$
$I_{1} = \dfrac{mr^{2}}{2} + M (2R + R)^{2} = 9.5MR^{2}$
$I_{2} = \dfrac{mr^{2}}{2} + MR^{2} = 2 MR^{2}$
$I = I_{1} + I_{2}$
$I = 9.5Mr^{2} + M R^{2}$
$I= 11.5MR^{2}$
Question 9
1 / -0

A body is thrown up with a velocity $29.23ms!$ $V$ distance travelled in last second of upward motion is

A
$2.3m$

B
$6m$

C
$9.8m$

D
$4.9m$

Solution

Total time of motion:
$v=u+at$
$x_0=29.23-9.8\times 6$
$xt=2.98\ sec$
distance total
$\Rightarrow v^2=u^2+2as$
$\Rightarrow 0=(29.23)^2=2\times 9.8 \times s$
$\Rightarrow s=43.59\ m$
Now in $1.96\ sec$
$s=29.3\times 1.98-\dfrac{1}{2}\times 9.8 (1.98)^2$
$s=36.6654$
in last $s= 43.59-38.66$
$=4.93m$
Question 10
1 / -0

A body is projected with velocity 'u' so that the maximum height is thrice the horizontal range. Then the maximum height is

A
$\dfrac{72u^2}{145\,g}$

B
$\dfrac{6}{\sqrt{145}}\dfrac{u^2}{g}$

C
$\dfrac{u^2}{2g}$

D
$\dfrac{145}{72}\dfrac{u^2}{g}$

Solution

$H_{max} = \dfrac{4^2sin^2\theta}{2g}$
Range $= \dfrac{4^2sin2\theta}{g} = \dfrac{4^2sin\theta\,cos\theta}{g}$
$\dfrac{H}{3} = R = \dfrac{4sin2\theta}{2 \times 3g} = \dfrac{4sin\theta\,cos\theta}{g}$
$tan\theta = \dfrac{12}{1}$
$sin\theta = \dfrac{12}{\sqrt{12^2 + 1^2}} = \dfrac{1}{\sqrt{145}}$
$H_{max} = \dfrac{4^2sin^2\theta}{2g} = \dfrac{4^2\left ( \dfrac{12}{\sqrt{145}} \right )^2}{2g}$
$= \dfrac{4^2 \times 144}{145 \times 2 \times g} = \dfrac{72u^2}{145\,g}$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Basic Science Test 11

Result

Basic Science Test 11

Score

-

Rank

-

SHARING IS CARING

Question 1

10m/s,010m/s,010m/s,0

60m/s,060m/s,060m/s,0

0,00,00,0

0,10m/s0,10m/s0,10m/s

Solution

Question 2

34.5N34.5N34.5N

22.8N22.8N22.8N

39.2N39.2N39.2N

57.6N57.6N57.6N

Solution

Question 3

Net force

Centripetal acceleration

Angular velocity

All of these

Solution

Question 4

10π10\pi10π

20π20\pi20π

30π30\pi30π

40π40\pi40π

Solution

Question 5

27tan⁡θ\frac{2}{7}\tan\theta72​tanθ

13gtan⁡θ\frac{1}{3}g\tan\theta31​gtanθ

12tan⁡θ\frac{1}{2}\tan\theta21​tanθ

25tan⁡θ\frac{2}{5}\tan\theta52​tanθ

Solution

Question 6

12m/s12 m/s12m/s

8m/s8 m/s8m/s

6m/s6 m/s6m/s

4m/s4 m/s4m/s

Question 7

5m5m5m

2m2m2m

000

none of these

Solution

Question 8

6.5mR2 6.5 m R^26.5mR2

12mR212m R^212mR2

6mR26 m R^26mR2

11.5mR211.5 m R^211.5mR2

Solution

Question 9

2.3m2.3m2.3m

6m6m6m

9.8m9.8m9.8m

4.9m4.9m4.9m

Solution

Question 10

72u2145 g\dfrac{72u^2}{145\,g}145g72u2​

6145u2g\dfrac{6}{\sqrt{145}}\dfrac{u^2}{g}145​6​gu2​

u22g\dfrac{u^2}{2g}2gu2​

14572u2g\dfrac{145}{72}\dfrac{u^2}{g}72145​gu2​

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$10m/s,0$

$60m/s,0$

$0,0$

$0,10m/s$

$34.5N$

$22.8N$

$39.2N$

$57.6N$

$10\pi$

$20\pi$

$30\pi$

$40\pi$

$\frac{2}{7}\tan\theta$

$\frac{1}{3}g\tan\theta$

$\frac{1}{2}\tan\theta$

$\frac{2}{5}\tan\theta$

$12 m/s$

$8 m/s$

$6 m/s$

$4 m/s$

$5m$

$2m$

$0$

$6.5 m R^2$

$12m R^2$

$6 m R^2$

$11.5 m R^2$

$2.3m$

$6m$

$9.8m$

$4.9m$

$\dfrac{72u^2}{145\,g}$

$\dfrac{6}{\sqrt{145}}\dfrac{u^2}{g}$

$\dfrac{u^2}{2g}$

$\dfrac{145}{72}\dfrac{u^2}{g}$