Basic Science Test 21

Result

Basic Science Test 21

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

The initial velocity of a particle is u (at t = 0 ) and the acceleration a is given by $$ at^{1/2}$$ Which of the following relation is valid ?

A
$$ v = un + at^{32} $$

B
$$ v = u + \dfrac{3 at^{3}}{2} $$

C
$$ v = u + \dfrac{2}{5} at^{1/2} $$

D
$$ v = u + at^{1/2} $$
Question 2
1 / -0

A particle of mass 40 g is moving in a circular orbit of radius 50 cm with frequengy 300 rpm. The work done in in increasing its frequency to 600 rpm is :

A
$$12.6 J$$

B
$$8.4 J$$

C
$$46.1 J$$

D
$$14.8 j$$
Question 3
1 / -0

An object measures $$3 \ cm$$ by $$9 \ cm$$ by $$4 \ cm$$ and weighs $$54 \ grams$$. If another object made from the same material measures $$6 \ cm$$ by $$2 \ cm$$ and $$3 \ cm$$, what would be the weight of the second object in terms of the first object?

A
3 times heavier

B
2 times heavier

C
The same weight

D
3 times lighter

Solution

According to question given
$$v_1= 3 \times 9 \times 4= 108$$
$$v_2 = 6 \times 2 \times 3= 36$$
$$\frac{v_1}{v_2} = \frac{w_1}{w_2}\implies w_1= 3 w_2$$
Option A
Question 4
1 / -0

What is the moment of inertia of ring about its diameter?

A
$$MR^2$$

B
$$\dfrac{MR^2}{2}$$

C
$$\dfrac{3}{4}MR^2$$

D
$$\dfrac{5}{4}MR^2$$

Solution

The moment of inertia of ring about its axis passing through the center is $$MR^2$$
$$I_Z = MR^2$$
Here $$I_Y = I_X = I$$
Now applying perpendicular axis theorem,
$$I_X + I_Y = I_Z^2$$
$$2 I = MR^2$$
$$I = \dfrac{MR^2}{2}$$
Question 5
1 / -0

A 1 kg particle strikes a wall with velocity 1 m/s at an angle $$30^{\circ}$$ and reflects at the same angle. If it remains in contact with wall for 0.1 s, then the force is

A
0

B
$$10\sqrt{3}$$ N

C
$$20\sqrt{3}$$ N

D
$$40\sqrt{3}$$ N

Solution

$$f = $$ Rate of change of momentum.

$$f = \dfrac{mV_{final} - mV_{initial}}{t} $$

$$f = \dfrac{2\times\dfrac{\sqrt{3}}{2}}{0.1}$$

$$f = 10\sqrt{3}$$ N
Question 6
1 / -0

An object moving horizontally with kinetic energy of 800 J experiences a constant opposing force of 100 N while moving from a to b (where ab = 2m ) . The energy of particle at is :

A
$$700 J$$

B
$$400 J$$

C
$$600 J$$

D
$$300 J$$

Solution

Answer
F = 100 N
d = 2m
energy of particle at b = ?
In passing 2m path against 200 N force particle will do work done & lose k .E
k.E lost = $$ F \times DD $$
= 200 N
Energy of particle a + b
= 800 - 200
= 600 J
Question 7
1 / -0

A stone is projected horizontally with a velocity 9.8 m/s from a tower of height 100m. Its velocity one second after projection is

A
$$9.8 ms^{-1}$$

B
$$4.9 ms^{-1}$$

C
$$9.8\sqrt{2} ms^{-1}$$

D
$$4.9\sqrt{2} ms^{-1}$$

Solution

For projectile at some height you will have to break the velocity into two components-
horizontal and vertical

The horizontal component is the initial velocity imparted; 9.8m/s. this component will remain unchanged throughout.

Squaring and adding this we get $$\sqrt{2}\times9.8 m/s$$ as final answer.
Question 8
1 / -0

Two discs have the same mass and thickness. Their materials are of densities $$d_1$$ and $$d_2.$$ The ratio of their moment of inertia about the central axis will be:

A
$$d_1d_2:1$$

B
$$d_1:d_2$$

C
$$d_2^2:d_1^2$$

D
$$1:d_1d_2$$

Solution

Given:-
$$M_1 = M_2$$
$$\implies d_1 \times Volume_1 = d_2 \times Volume_2$$
$$\implies d_1 \times \pi D_1 \times thickness_1 = d_2 \times \pi D_2 \times thickness_2$$
$$\implies \dfrac{R_1}{R_2} = \dfrac{D_1}{D_2} = \dfrac{d_1}{d_2}$$
Now,
$$\implies \dfrac{I_1}{I_2} = \dfrac{\dfrac{M R_1 ^2}{2}}{\dfrac{MR_2 ^2}{2}}= \left( \dfrac{d_2}{d_1} \right)^2$$
Question 9
1 / -0

If uncertainty in the measurement of position and momentum are equal . What is uncertainty in the measurement of velocity

A
$$\dfrac{1}{2} \sqrt{\dfrac{mh}{\pi}}$$

B
$$\dfrac{1}{2 \pi} \sqrt{\dfrac{m}{h}}$$

C
$$\dfrac{1}{2m} \sqrt{\dfrac{h}{\pi}}$$

D
None of these

Solution

$$ \Delta b = \Delta R $$

$$ \Delta b \times \Delta K \geq \dfrac{h}{4\pi} $$

$$ \Rightarrow (\Delta b)^{2} \geq \dfrac{h}{4\pi} $$

$$ \Rightarrow (m \Delta v) ^{2} \geq \dfrac{h}{4\pi} $$

$$ \Rightarrow \Delta V \geq \sqrt{\dfrac{h}{4 \pi m ^{2}}} $$

$$ \geq \dfrac{1}{2m } \sqrt{\dfrac{h}{\pi}}$$
Question 10
1 / -0

A wheel of moment of inertial $$2 Kg m^2$$ is rotating about an axis passing through centre and perpendicular to its plane at a speed 60 rad/s. Due to friction, it comes to rest in 5 minutes. The angular momentum of the wheel three minutes before it stops rotating is

A
$$24 Kg m^2/s$$

B
$$48 Kg m^2/s$$

C
$$72 Kg m^2/s$$

D
$$96 Kg m^2/s$$

Solution

Given:-

$$I= 2Kgm^2$$

$$w_{initial} = 60rad/s$$

At time$$(t=5 \ min)$$ $$w_{final} = 0$$

We have to find $$w_{(t= 2 \ min)} $$

$$w_{final} = 0= w_{initial} - \alpha t \Rightarrow \alpha= \dfrac{w_{initial} - w_{final} }{t} = \dfrac{60-0}{5} = 12 rad/s^2$$

Now $$w_{(t= 2 \ min)} = 60 - 12\times 2 = 36 rad/s$$

$$\therefore L = I\omega = 2\times36 = 72 Kg m^2/s$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Basic Science Test 21

Result

Basic Science Test 21

Score

-

Rank

-

SHARING IS CARING

Question 1

$$ v = un + at^{32} $$

$$ v = u + \dfrac{3 at^{3}}{2} $$

$$ v = u + \dfrac{2}{5} at^{1/2} $$

$$ v = u + at^{1/2} $$

Question 2

$$12.6 J$$

$$8.4 J$$

$$46.1 J$$

$$14.8 j$$

Question 3

3 times heavier

2 times heavier

The same weight

3 times lighter

Solution

Question 4

$$MR^2$$

$$\dfrac{MR^2}{2}$$

$$\dfrac{3}{4}MR^2$$

$$\dfrac{5}{4}MR^2$$

Solution

Question 5

0

$$10\sqrt{3}$$ N

$$20\sqrt{3}$$ N

$$40\sqrt{3}$$ N

Solution

Question 6

$$700 J$$

$$400 J$$

$$600 J$$

$$300 J$$

Solution

Question 7

$$9.8 ms^{-1}$$

$$4.9 ms^{-1}$$

$$9.8\sqrt{2} ms^{-1}$$

$$4.9\sqrt{2} ms^{-1}$$

Solution

Question 8

$$d_1d_2:1$$

$$d_1:d_2$$

$$d_2^2:d_1^2$$

$$1:d_1d_2$$

Solution

Question 9

$$\dfrac{1}{2} \sqrt{\dfrac{mh}{\pi}}$$

$$\dfrac{1}{2 \pi} \sqrt{\dfrac{m}{h}}$$

$$\dfrac{1}{2m} \sqrt{\dfrac{h}{\pi}}$$

None of these

Solution

Question 10

$$24 Kg m^2/s$$

$$48 Kg m^2/s$$

$$72 Kg m^2/s$$

$$96 Kg m^2/s$$

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.