Physics Test

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Physics Test - 15

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Question 1
4 / -1

In the circuit given below, an A.C. voltage V = 220 volt (r.m.s.) is applied. The maximum voltage across the capacitor will be

A
220 $\times \sqrt{2}$ volts

B
110 $\times \sqrt{2}$ volts

C
220 volts

D
Cannot be determined

Solution

The maximum voltage across the capacitor will be,

V_max = $V_{r . m . s} \times \sqrt{2} = 220 \times \sqrt{2}$
Question 2
4 / -1

If M0 is the mass of an oxygen isotope 8O17, and Mp and Mn are the masses of a proton and a neutron, respectively, then what is the nuclear binding energy of the isotope

A
(M0- 8M)c2

B
(8Mp+ 9Mn- Mo)c2

C
Moc2

D
(Mo- 17Mn)c2

Solution

Here, Z = 8 and A = 17, therefore,
B.E. = (8M_p+ (17 - 8)M_n- M_o)c²
= (8M_p+ 9M_n- M_o)c²
Question 3
4 / -1

An electron (q = 1.6 x 10-19C) is moving at right angle to the uniform magnetic field of 3.534 x 10-5T. The time taken by the electron to complete a circular orbit is

A
2 µs

B
4 µs

C
3 µs

D
1 µs

Solution

Key Idea When $a$ charged particle is projected perpendicular to a magnetic field its path is circular in a plane perpendicular to the field. When the charged particle is moving perpendicular to the field, in this situation, as the angle between $\overrightarrow{\mathbf{B}}$ and $\overrightarrow{\mathbf{v}}$ is $\theta=90^{\circ},$ so the force will be maximum $(=q v B)$ and always perpendicular to motion (and also field), so the path will be a circle (with its plane perpendicular to the field) as in a circle tangent and radius are always perpendicular to each other. As here centripetal force is provided by the force $q v B$
\[
\frac{m v^{2}}{r}=q v B
\]
ie,
\[
r=\frac{m v}{q B}
\]

As in uniform circular motion $v=r \omega,$ so the angular frequency of circular motion will be given by
\[
\omega=\frac{v}{r}=\frac{q B}{m} \quad \text { [Using }
\]
and hence the time period
\[
T=\frac{2 \pi}{\omega}=\frac{2 \pi m}{q B}
\]
Given, $B=3.534 \times 10^{-5} \mathrm{T}$
$q=1.6 \times 10^{-19} \mathrm{C}, \quad m=9.1 \times 10^{-31} \mathrm{kg}, T=?$
From Eq. (ii), we get
\[
\begin{aligned}
\therefore T &=\frac{2 \times 3.14 \times 9.1 \times 10^{-31}}{3.534 \times 10^{-5} \times 1.6 \times 10^{-19}} \\
&=1 \times 10^{-6} \mathrm{s} \\
&=1 \mu \mathrm{s}
\end{aligned}
\]
Note This principle has been used in a large number of devices such as cyclotron, bubble chamber (a particle detector) or mass spectrometer etc.
Question 4
4 / -1

In a stationary lift, a man is standing with a bucket full of water, having a hole at its bottom. The rate of flow of water through this hole is R0. If the lift starts to move up and down with the same acceleration such that the water flows at the rate of Ru and Rd, then

A
R0> Ru> Rd

B
Ru> R0> Rd

C
Rd> R0> Ru

D
Ru> Rd> R0

Solution

Rate of flow of water is proportional to effective force pressure on water.
When the lift moves upwards, $R_{U} \alpha m(g+a)$.
When the lift moves downwards, $\mathrm{R}_{\mathrm{d}} \alpha \mathrm{m}(\mathrm{g}-\mathrm{a})$
When the lift is stationary, $\mathrm{R}_{0} \alpha \mathrm{mg}$
Therefore, $R_{\mathrm{U}}>R_{O}>R_{\mathrm{d}}$
Question 5
4 / -1

P and Q are like parallel forces. If P is moved parallel to itself through a distance x, then the resultant of P and Q moves through a distance:

A
$\frac{P_{X}}{p+Q}$

B
$\frac{P_{X}}{p-Q}$

C
$\frac{P x}{p+2 Q}$

D
none of these

Solution

Ist case. P.AC $=$ Q. $C B$

$2^{\text {nd }}$ case, $\mathrm{P}$ is shifted by a distance $\mathrm{AA}^{2}=\mathrm{x}$ and as a consequence suppose the resultant $\mathrm{R}$ shifts from
C to $C^{\circ}$ where $C C^{\circ}=y$
$\therefore \quad P_{.} A^{*} C^{2}=Q . C^{*} B$
Or $P(A C-x+y)=Q(C B-y)$
Or $P(-x+y)=-Q y$
$\because \quad P_{.} A . C=Q . C B \quad$ by (1)
$(P+Q) y=P x$ or $y=\frac{P x}{P+Q}$
Question 6
4 / -1

A small circular flexible loop of wire of radius r carries a current l. It is placed in a uniform magnetic field B. The tension in the loop will be doubled if :

A
l is doubled

B
B is halved

C
r is doubled

D
either B is doubled or l is doubled

Solution

Consider a small arc of the length of the circular loop, force acting on it due to the magnetic field.
$\mathrm{dF}=\mathrm{B}$ ld $,$, If $\mathrm{T}$ is the tension in the string, then
$2 \mathrm{T} \sin \theta=\mathrm{B}|\mathrm{d}| \Rightarrow \mathrm{T}=\mathrm{B} \mathrm{IR} \quad(\cup \operatorname{sing} \mathrm{dl}=\mathrm{R} \mathrm{d} \theta$ and $\sin \theta \approx \theta)$
Hence Tension in the string will get double if either current is doubled or Magnetic field is doubled.
Question 7
4 / -1

A triangle $A B C$ forces $\vec{P}, \overrightarrow{\mathrm{Q}}, \overrightarrow{\mathrm{R}}$ acting along IA. IB and $\mathrm{IC}$, respectively, in equilibrium, where I is the incentre of ABC. Then, $\overrightarrow{\mathrm{P}}: \overrightarrow{\mathrm{Q}} \cdot \overrightarrow{\mathrm{R}}$ is

A
cos A : cos B : cos C

B
$\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$

C
$\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}$

D
sin A : sin B : sin C

Solution

Using Lami's Theorem $\therefore P: Q: R=\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}$
Question 8
4 / -1

The resultant of P and Q is R. If P is reversed, Q remains the same, the new resultant S is at right angle to R. The value of P / Q is

A
1

B
2

C
1/2

D
none of these

Solution

Correct Answer: 1 $\vec{R}=\vec{P}+\vec{Q} \ldots \ldots(i)$
$\vec{S}=-\vec{P}+\vec{Q} \ldots \ldots(i i)$
Adding
$(i)$ and $(i i)$ we get $\vec{R}+\vec{S}=2 \vec{Q} \ldots . .(i i i)$
subtracting
$(i)$ and $(i i)$ we get $\vec{R}-\vec{S}=2 \vec{P} \ldots . .(i v)$
Divide eq (iii) and eq (iv) $\frac{2 \vec{P}}{2 \vec{Q}}=\frac{\vec{R}-\vec{S}}{\vec{R}+\vec{S}}=\frac{\vec{R} \cdot \vec{R}-(\vec{S} \cdot \vec{R})}{\vec{R} \cdot \vec{R}+(\vec{S} \cdot \vec{R})}=1$
$\Rightarrow \frac{\vec{P}}{\vec{Q}}=1$
Question 9
4 / -1

A horizontal pipe of cross-sectional diameter 5 cm carries water at a velocity of 4 m/s. The pipe is connected to a smaller pipe with a cross-sectional diameter 4 cm. What is the velocity of water through the smaller pipe?

A
6.25 m/s

B
5.0 m/s

C
3.2 m/s

D
2.56 m/s

Solution

From equation of continuity, $a_{1} v_{1}=a_{2} v_{2}$
\[
\pi(2.5)^{2} \times 4=\pi(2)^{2} \times v_{2}
\]
$v_{2}=6.25 \mathrm{m} / \mathrm{s}$
Question 10
4 / -1

The cylindrical tube of a spray pump has a cross-section of 8 cm2, one end of which has 40 fine holes, each of area 10-8m2. If the liquid flows inside the tube with a speed of 0.15 m min-1, then the speed with which the liquid is ejected through the holes is

A
50 ms-1

B
5 ms-1

C
0.05 ms-1

D
0.5 ms-1

Solution

According to equation of continuity, $a v=$ constant
$\therefore$ For tube, $\left(8 \times 10^{-4}\right) \times\left(\frac{0.15}{60}\right)=a_{1} v_{1}$
For holes, $\left(40 \times 10^{-8}\right) \times v=a_{2} v_{2}$
$\begin{aligned} \therefore \quad a_{2} v_{2} &=a_{1} v_{1} \\ \therefore \quad 40 \times 10^{-8} \times v &=\frac{8 \times 10^{-4} \times 0.15}{60} \\ \Rightarrow \quad v &=\frac{8 \times 10^{-4} \times 0.15}{40 \times 10^{-8} \times 60} \\ &=5 \mathrm{m} / \mathrm{s} \end{aligned}$

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Physics Test - 15

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Physics Test - 15

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Question 1

220×2volts

110×2volts

220 volts

Cannot be determined

Solution

Question 2

(M0- 8M)c2

(8Mp+ 9Mn- Mo)c2

Moc2

(Mo- 17Mn)c2

Solution

Question 3

2 µs

4 µs

3 µs

1 µs

Solution

Question 4

R0> Ru> Rd

Ru> R0> Rd

Rd> R0> Ru

Ru> Rd> R0

Solution

Question 5

\(\frac{P_{X}}{p+Q}\)

\(\frac{P_{X}}{p-Q}\)

\(\frac{P x}{p+2 Q}\)

none of these

Solution

Question 6

l is doubled

B is halved

r is doubled

either B is doubled or l is doubled

Solution

Question 7

cos A : cos B : cos C

\(\cos \frac{A}{2}: \cos \frac{B}{2}: \cos \frac{C}{2}\)

\(\sin \frac{A}{2}: \sin \frac{B}{2}: \sin \frac{C}{2}\)

sin A : sin B : sin C

Solution

Question 8

1

2

1/2

none of these

Solution

Question 9

6.25 m/s

5.0 m/s

3.2 m/s

2.56 m/s

Solution

Question 10

50 ms-1

5 ms-1

0.05 ms-1

0.5 ms-1

Solution

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220 $\times \sqrt{2}$ volts

110 $\times \sqrt{2}$ volts