Physics Test

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Physics Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Two wires of same material and the same diameter have length in the ratio 2 : 5. They are stretched by the same force. The ratio of work done in stretching them is

A
5 : 2

B
2 : 5

C
2 : 3

D
3 : 1

Solution

Work Done $=\frac{1}{2} \times $ Force $\ \times $ Elongation
$Y=\frac{F l}{A \triangle l}$
$\Delta l=\frac{F l}{A Y}$
Work Done$=\frac{1}{2} \times F \times \frac{F l}{A Y}=\frac{F^{2} L}{2 A Y}$
since, the Force, Area and young's modulus is same,
$\therefore \quad$ Work Done $\propto \quad$ length
$\frac{W_{1}}{W_{2}}=\frac{l_{1}}{l_{2}}=\frac{2}{5}$
Question 2
1 / -0

There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of the wire is:

A
0.50

B
– 0.50

C
0.25

D
– 0.25

Solution

Let the material of length $l$ and side $s$
If a material maintains constant volume during stretching $V=l \times s^{2}$
Differentiate wrt $d l$
$d V=s^{2} \cdot d l+l .2 s . d s$
$d l . s=2 l . d s$
$\frac{d s}{d l}=-\frac{1}{2} \frac{s}{l}$
$\nu=-\frac{\frac{d s}{s}}{\frac{d l}{l}}=-\frac{1}{2}$
Question 3
1 / -0

If in a wire of Young's modulus 'Y', longitudinal strain 'X' is produced, then the potential energy stored in its unit volume will be

A
0.5 Y X²

B
0.5 Y² X

C
2 Y X²

D
Y X²

Solution

Potential energy stored per unit volume of a wire $\frac{1}{2}$stress$\times$strain
$Y=\frac{\text {stress}}{\text {strain}}$
stress$=Y \times$strain
$U=\frac{1}{2} \times Y X \times X=\frac{1}{2} Y X^{2}$
$={0.5} Y X^{2}$
Question 4
1 / -0

The energy stored per unit volume in a copper wire which produces a longitudinal strain of 0.1% is (Y = 1.1 x 10¹¹ N/m²)

A
11 × 103 J/m³

B
5.5 × 10³ J/m³

C
5.5 × 10⁴ J/m³

D
7 × 10⁴ J/m³

Solution

When a wire is stretched, some work is done against the internal restoring forces, acting between particles of the wire. This work done appears as elastic potential energy in the wire.
Elastic potential energy $(U)$ is given by
\[
\begin{aligned}
U &=\frac{1}{2} F \times \Delta l \\
&=\frac{1}{2} \times \frac{F}{a} \times \frac{\Delta l}{l} \times a l
\end{aligned}
\]
where $l$ is length of wire, $a$ is area of cross section of wire, $F$ is stretching force and $\Delta l$ is increase in length. Equation (i), may be written as $U=\frac{1}{2} \times$ stress $\times$ strain $\times$ volume of the wire
$\therefore$ Elastic potential energy per unit volume of the wire
\[
\begin{array}{l}
u=\frac{U}{a l}=\frac{1}{2} \times \text { stress } \times \text { strain } \\
=\frac{1}{2} \times(\text { Young's modulus } \times \text { strain }) \times \text { strain } \\
=\frac{1}{2} \times(Y) \times(\text { strain })^{2} \\
\text { Hence, } u=\frac{1}{2} \times 1.1 \times 10^{11} \times\left(\frac{0.1}{100}\right)^{2} \\
=5.5 \times 10^{4} \mathrm{J} / \mathrm{m}^{3}
\end{array}
\]
Question 5
1 / -0

Directions: The following question has four choices out of which ONLY ONE is correct.
Which of the following is not a unit of Young's modulus?

A
Nm^–1

B
Nm^–2

C
dyne cm^–2

D
MPa

Solution

Young's modulus $Y=\frac{\text {Stress}}{\text {Strain}}$
As we know, strain is a dimensionless quantity whereas stress is defined as force per unit area. Stress has units as $N m^{-2}$ (Sl unit), Dyne $\mathrm{cm}^{-2}$ (cgs unit) or Pascal $(P a)$ As Young's modulus has same unit as that of stress, thus $N m^{-1}$ is not a unit of Young's modulus.
Question 6
1 / -0

A needle made of bismuth is suspended freely in a magnetic field. The angle which the needle makes with the magnetic field is

A
0⁰

B
45⁰

C
90⁰

D
180⁰

Solution

Bismuth is a diamagnetic substance, so when placed in an external magnetic field it rotates such that its axis becomes perpendicular to the magnetic field.
Question 7
1 / -0

The magnetic susceptibility of a material of a rod is 449. Permeability of vacuum is 4π x 10^-7 henry/metre. The absolute permeability of the material of the rod (in henry/metre) is

A
π x 10^-4

B
2π x 10^-4

C
3π x 10^-4

D
4π x 10^-4

Solution

Given:
$\mu_{o}=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}$
Magnetic susceptibility of a material of rod:
$\chi_{m}=499$
$\therefore$ Absolute permeability of material of rod:
$\mu=\mu_{o}\left(1+\chi_{m}\right)$
or, $\quad \mu=4 \pi \times 10^{-7}(1+499)=2 \pi \times 10^{-4} \mathrm{Hm}^{-1}$
Question 8
1 / -0

A magnet of length 10 cm and magnetic moment 1 Am² is placed along the side AB of an equilateral triangle ABC. If the length of side AB is 10 cm, then the magnetic field at point C is

A
10^-9 T

B
10^-7 T

C
10^-5 T

D
10^-4 T

Solution

For a bar magnet, magnetic induction at a point on its axis is given by the formula,
$B=\frac{\mu_{0}}{4 \pi} \frac{M}{\left(d^{2}+l^{2}\right)^{3 / 2}}$
$\frac{\mu_{0}}{4 \pi}=10^{-7}$
Where $l$ is the length of bar magnet and $d$ is distance of point from the centre of bar magnet. Hence, here, $l=0.1 m$ and $d=0.05 \sqrt{3} m$
Putting in the formula,
we get
$B=10^{-4} T$
Question 9
1 / -0

A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75⁰. One of the fields has a magnitude of $\sqrt{2} \times$ 10^-2 T. The dipole attains stable equilibrium at an angle of 30⁰ with this field. What is the magnitude of the other field?

A
0.01 T

B
0.02 T

C
0.03 T

D
0.04 T

Solution

Torque on a dipole is given by: $\vec{\tau}=\vec{M} \times \vec{B}$ where: $M:$ Magnetic moment, $B$ : Magnetic flux density
It is given that the dipole is in equilibrium. Hence, the configuration of the magnetic field vectors and the magnetic moment is as shown in the figure so that the torque due to the two fields are opposite in direction.
$\tau_{1}=\tau_{2}$
$M B_{1} \sin \left(30^{\circ}\right)=M B_{2} \sin \left(45^{\circ}\right)$
$B_{2}=\frac{\sqrt{2}\times10^{-2} \times \frac{1}{2}}{\frac{1}{\sqrt{2}}}$
$B_{2}=0.01 \mathrm{T} $
Question 10
1 / -0

A steel wire of length 'l' has a magnetic moment M. It is bent into a semicircular arc. The new magnetic moment is

A
M

B
2M/π

C
M/π

D
Mπ

Solution

When wire is bent in the form of semicircular arc then, $l=\pi r$
$\therefore$ The radius of semicircular arc, $r=l / \pi$
Distance between two end points of semicircular wire $=2 r=\frac{2 l}{\pi}$
$\therefore$ Magnetic moment of semicircular wire $=m \times 2 r=m \times \frac{2 l}{\pi}=\frac{2}{\pi} m l$
But $m l$ is the magnetic moment of straight wire
i.e., $m l=M$
$\therefore$ New magnetic moment $=\frac{2}{\pi} M$

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Re-Attempt Weekly Quiz Competition

Physics Test - 15

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Physics Test - 15

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SHARING IS CARING

Question 1

5 : 2

2 : 5

2 : 3

3 : 1

Solution

Question 2

0.50

– 0.50

0.25

– 0.25

Solution

Question 3

0.5 Y X2

0.5 Y2 X

2 Y X2

Y X2

Solution

Question 4

11 × 103 J/m3

5.5 × 103 J/m3

5.5 × 104 J/m3

7 × 104 J/m3

Solution

Question 5

Nm–1

Nm–2

dyne cm–2

MPa

Solution

Question 6

00

450

900

1800

Solution

Question 7

π x 10-4

2π x 10-4

3π x 10-4

4π x 10-4

Solution

Question 8

10-9 T

10-7 T

10-5 T

10-4 T

Solution

Question 9

0.01 T

0.02 T

0.03 T

0.04 T

Solution

Question 10

M

2M/π

M/π

Mπ

Solution

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0.5 Y X²

0.5 Y² X

2 Y X²

Y X²

11 × 103 J/m³

5.5 × 10³ J/m³

5.5 × 10⁴ J/m³

7 × 10⁴ J/m³

Nm^–1

Nm^–2

dyne cm^–2

0⁰

45⁰

90⁰

180⁰

π x 10^-4

2π x 10^-4

3π x 10^-4

4π x 10^-4

10^-9 T

10^-7 T

10^-5 T

10^-4 T