Physics Test

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Physics Test - 6

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Question 1/10
4 / -1

Three identical spheres, each of mass 1 kg, are kept as shown in the figure below, touching each other with their centres on a straight line. If their centres are marked respectively as P, Q and R, then the distance of the centre of mass of the system from P is

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A
$\frac{P Q + P R + Q R}{3}$

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B
PQ

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C
$\frac{P Q + Q R}{3}$

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D
$\frac{P R + O R}{2}$

Solutions

As the spheres are uniform, the mass is equally distributed about Q along x-axis; centre of mass is $Q$ ; distance of the centre of mass is $PQ$ . $CM$ can be calculated as:
First sphere is centred at origin $(x = 0)$ , second sphere is centred at $x = P Q$ , third sphere is centred at $x = PR$ . $PR = 2 PQ$
$x_{c m} = \frac{(m \times 0) + (m \times P Q) + (m \times 2 P Q)}{3 m} = \frac{3 m \times P Q}{3 m} = P Q$
Question 2/10
4 / -1

A horizontal platform is rotating with uniform angular velocity around the vertical axis passing through its centre. At some instant of time, a viscous fluid of mass 'm' is dropped at the centre and is allowed to spread out and finally fall. The angular velocity during this period-

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A
Decreases continuously

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B
Decreases initially and increases again

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C
Remains unaltered

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D
Increases continuously

Solutions

According to law of conservation of momentum,
l_ω = constant
When viscous fluid of mass m is dropped and start spreading out then its moment of inertia increases and angular velocity decreases. But when it start falling then its moment of inertia again starts decreasing and angular velocity increases.
Question 3/10
4 / -1

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

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A
M

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B
2M/π

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C
M/π

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D
Mπ

Solutions

If $m$ is pole strength, then $m = \frac{M}{l}$
When the wire is bent into a semicircular arc, the separation between the two poles
changes from I to $2 r$ , where $r$ is radius of the semicircular arc.
since $I = π r$ or $r = 1 / π,$ the new magnetic moment of the stell wire,
$M^{'} = m \times 2 r = \frac{M}{l} \times \frac{2 l}{π} = \frac{2 M}{π}$
Question 4/10
4 / -1

A coil takes a current of 2 A and 200 W power from an alternating current source of 220 V, 50 Hz. What is the inductance of the coil?

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A
0.312 H

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B
0.142 H

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C
0.285 H

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D
1.07 H

Solutions

$R = \frac{P}{I^{2}}$
$= \frac{200}{4} = 50 Ω$
$Z = \sqrt{R^{2} + X_{L}^{2}} = \sqrt{(50)^{2 + X_{L}^{2}}}$
$Z = V / I = 220 / 2 = 110 ohm$
$X_{L}^{2} = 12100 - 2500 = 9600$
$X_{L} = 98 ohm$
$L = \frac{X_{L}}{ω} = \frac{X_{L}}{2 π f} = \frac{98}{2 \times 3.14 \times 50} = 0.312 H$
Question 5/10
4 / -1

A convex lens made up of a material of refractive index μ₁ is immersed in a medium of refractive index μ₂ as shown in the figure. The relation between μ₁ and μ₂ is

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A
μ₁ < μ₂

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B
μ₁ > μ₂

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C
μ₁ = μ₂

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D
$μ_{1} = \sqrt{μ_{2}}$

Solutions

In a medium of higher refractive index, a convex lens diverges a parallel beam of light, thus behaving a diverging lens. Hence, $μ_{1} < μ_{2}$
Question 6/10
4 / -1

Three particles, each of mass 'm' g, are situated at the vertices of an equilateral triangle ABC of side l cm (as shown in the figure). The moment of inertia of the system about a line AX perpendicular to AB in the plane of ABC in gram-cm² unit will be

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A
(3/4) ml²

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B
2 ml²

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C
(5/4) ml²

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D
(3/2) ml²

Solutions

The moment of inertia of the

system $= m_{A} r_{A}^{2} + m_{B} r_{B}^{2} + m_{C} r_{C}^{2}$
$= m_{A} (0)^{2} + m (l)^{2} + m {(l \sin 30^{\circ})}^{2}$
$= m l^{2} + m l^{2} \times (1 / 4) = (5 / 4) m l^{2}$
Question 7/10
4 / -1

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

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A
0°

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B
45°

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C
90°

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D
180°

Solutions

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 8/10
4 / -1

A stone is dropped from a certain height which can reach the ground in $5 s$ . If the stone is stopped after 3 s of its fall and then allowed to fall again, then the time taken by the stone to reach the ground for the remaining distance is:

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A
2 s

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B
3 s

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C
4 s

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D
8 s

Solutions

Let $h$ be the height and $t = 5 s$ .
Equations of motion can be written as:
$∴ h = \frac{1}{2} {gt}^{2} = \frac{1}{2} (9.8) (25) = 122.5 m$
Distance covered in $3 \sec = \frac{1}{2} (9.8) (3)^{2}$
$= 44.1 m$
Remaining distance $= 122.5 - 44.1 = 78.4 m$
If $t s$ is the required time, then:
$78.4 = 0 + \frac{1}{2} {gt}^{2}$
$\Rightarrow t^{2} = \frac{78.4 \times 2}{9.8}$
$\Rightarrow t^{2} = 16$
$\Rightarrow t = 4 s$
Question 9/10
4 / -1

The binding energies for nuclei ₁H¹, ₂He⁴, ₂₆Fe⁵⁶and ₉₂U²³⁵ are 2.22, 28.3, 492 and 17.86 MeV, respectively. The most stable nucleus is

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₁H¹

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B
₂He⁴

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C
₂₆Fe⁵⁶

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D
₉₂U²³⁵

Solutions

As the binding energy of Fe⁵⁶ is highest, so the most stable nucleus will be₂₆Fe⁵⁶.
Question 10/10
4 / -1

Sodium surface is illuminated by ultraviolet and visible radiations, successively and the stopping potential is determined. This stopping potential is

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A
equal in both cases

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B
more with ultraviolet light

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C
more with visible light

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D
varies randomly

Solutions

Ultraviolet light has lower wavelength, thus higher frequency.
We know, $h_{v} = ϕ + E, v =$ Frequency of radiation, $ϕ =$ Work function, $E =$ Maximum
energy of
released electrons
Stopping potential= $E / e, (e =$ charge of an electron) As frequency increases, E increases, and thus, stopping potential
increases.

Question 1/10

PQ+PR+QR3

PQ

PQ+QR3

PR+OR2

Question 2/10

Decreases continuously

Decreases initially and increases again

Remains unaltered

Increases continuously

Question 3/10

M

2M/π

M/π

Mπ

Question 4/10

0.312 H

0.142 H

0.285 H

1.07 H

Question 5/10

μ1 < μ2

μ1 > μ2

μ1 = μ2

μ1=μ2

Question 6/10

(3/4) ml2

2 ml2

(5/4) ml2

(3/2) ml2

Question 7/10

0°

45°

90°

180°

Question 8/10

2 s

3 s

4 s

8 s

Question 9/10

1H1

2He4

26Fe56

92U235

Question 10/10

equal in both cases

more with ultraviolet light

more with visible light

varies randomly

$\frac{P Q + P R + Q R}{3}$

$\frac{P Q + Q R}{3}$

$\frac{P R + O R}{2}$

μ₁ < μ₂

μ₁ > μ₂

μ₁ = μ₂

$μ_{1} = \sqrt{μ_{2}}$

(3/4) ml²

2 ml²

(5/4) ml²

(3/2) ml²

₁H¹

₂He⁴

₂₆Fe⁵⁶

₉₂U²³⁵