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Physics Test - 27

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Question 1
3 / -1

The displacement y of a particle executing periodical motion is given by y = 4 cos² (t/2) sin (1000t). This expression may be considered to be a result of the superposition of how many independent harmonic motions?

A
$5$

B
$4$

C
$3$

D
$2$

Solution

Hence above expression involves three different waves the displacement given is a result of superposition of three different waves.
Question 2
3 / -1

Three particles, each of mass $' m' g$ , are situated at the vertices of an equilateral triangle $A B C$ of side $l c m$ (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ in the plane of $A B C$ in gram- $c m^{2}$ unit will be:

A

B

C

D

Solution
Question 3
3 / -1

In an LCR series circuit, the potential difference between the terminals of the inductance is $60 V$ , that between the terminals of the capacitor is $30 V$ and that across the resistance is $40 V$ . Then, the supply voltage will be:

A
$50 V$

B
$70 V$

C
$130 V$

D
$10 V$

Solution

Given,

$V_{R} = 40 V$

$V_{L} = 60 V$

$V_{C} = 30 V$

For the series LCR circuit, the effective voltage is $V = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}}$

$V = \sqrt{{(40)}^{2} + {(60 - 30)}^{2}} = \sqrt{1600 + 900}$

$V = \sqrt{2500} = 50 V$
Question 4
3 / -1

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
$\frac{2 M}{π}$

C
$\frac{M}{π}$

D
$M π$

Solution

If m is a pole strength, then m = M/1

When the wire is bent into a semicircular arc, the separation between the two poles changes from l to 2r, where r is the radius of the semicircular arc.
Question 5
3 / -1

Two weights $' w_{1}'$ and $' w_{2}'$ are suspended by two strings on a frictionless pulley. When the pulley is pulled up with an acceleration $' g'$ , then the tension in the string is:

A
$\frac{4 W_{1} W_{2}}{W_{1} + W_{2}}$

B
$\frac{W_{1} W_{2}}{W_{1} + W_{2}}$

C
$\frac{2 W_{1} W_{2}}{W_{1} + W_{2}}$

D
$\frac{W_{1} + W_{2}}{2}$

Solution

Free body diagram,

$2 m_{1} g - T = m_{1} a$ ...(i)

$T - 2 m_{2} g = m_{2} a$ ...(ii)

From equation (i)

$a = \frac{2 m_{1} g - T}{m_{1}}$

Substituting value in equation (ii)

$T - 2 m_{2} g = m_{2} (\frac{2 m_{1} g - T}{m_{1}})$

or $T m_{1} - 2 m_{1} m_{2} g = 2 m_{2} m_{1} g - T m_{2}$

or $T (m_{1} + m_{2}) = 4 m_{2} m_{1} g$

or $T = \frac{4 m_{2} m_{1} g \times g}{(m_{1} + m_{2}) \times g} = \frac{4 W_{1} W_{2}}{(W_{1} + W_{2})}$
Question 6
3 / -1

A stone is dropped from a certain height which can reach the ground in $5$ seconds. If the stone is stopped after $3$ seconds 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-

A
$2$ seconds

B
$3$ seconds

C
$4$ seconds

D
$5$ seconds

Solution

Let $h$ be the height.

Time $=$$5$ seconds
$\therefore h=\frac{1}{2} g t^{2}=\frac{1}{2}(9.8)(25)=122.5 {~m}$

Distance covered in $3$ seconds=$\frac{1}{2}(9.8)(3)^{2}$
$=44.1 {~m}$

Remaining distance $=122.5-44.1=78.4 {~m}$
If $t$ seconds is the required time, then
$78.4=0+\frac{1}{2} g t^{2}$
$\Rightarrow t^{2}=\frac{784}{49}=16$
$\Rightarrow t=4$ seconds
Question 7
3 / -1

The binding energies for nuclei ${}_{1}H^{1}, {}_{2}H e^{4}, {}_{26}F e^{56}$ and ${}_{92}U^{235}$ are $2.22, 28.3, 492$ and $17.86 M e V$ , respectively. The most stable nucleus is:

A
${}_{1}H^{1}$

B
${}_{2}H e^{4}$

C
${}_{26}F e^{56}$

D
${}_{92}U^{235}$

Solution

As we know stability $\propto$ binding energy per nucleon.

${}_{1}H^{2} : \frac{2.22}{2} = 1.11 M e V$

${}_{2}H^{4} : \frac{28.3}{4} = 7.07 M e V$

${}_{26}F e^{56} : \frac{492}{56} = 8.78 M e V$

${}_{92}U^{235} : \frac{1786}{235} = 7.6 M e V$

So, by this ${}_{26}F e^{56}$ is the most stable nucleus as its binding energy per nucleon is the largest among the given four entities.
Question 8
3 / -1

$F_{1}$ and $F_{2}$ are focal lengths of the objective and the eyepiece of a telescope, respectively. The angular magnification for this telescope is equal to:

A
$\frac{F_{1}}{F_{2}}$

B
$\frac{F_{2}}{F_{1}}$

C
$\frac{F_{1} F_{2}}{F_{1} + F_{2}}$

D
$\frac{F_{1} + F_{2}}{F_{1} F_{2}}$

Solution

Angular magnification is the ratio of the focal length of the objective to the focal length of the eyepiece.

$∴ M = \frac{F_{1}}{F_{2}}$
Question 9
3 / -1

A horizontal platform is rotating with a uniform angular velocity around the vertical axis passing through its centre. At some instant in time, 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:

A
decreases continuously

B
decreases initially and increases again

C
remains unaltered

D
increases continuously

Solution

According to the law of conservation of momentum,

$l_{ω} =$ constant

When viscous fluid of mass $m$ is dropped and starts spreading out then its moment of inertia increases and angular velocity decreases. But when it starts falling then its moment of inertia again starts decreasing and angular velocity increases.
Question 10
3 / -1

For a certain radioactive substance, it is observed that after $4$ hours, only $6.25 %$ of the original sample is left undecayed. Which of the following statements is incorrect?

A
The half-life of the sample is $1$ hr.

B
The mean life of the sample is $\frac{1}{2}$ hr.

C
The decay constant of the sample is $(\ln 2) h r^{- 1}$ .

D
After further $4$ hours, the amount of the substance leftover would be only $0.39 %$ of the original amount.

Solution

We have

$6.25 % = \frac{6.25}{100} = \frac{1}{16} = \frac{1}{{(2)}^{4}}$

Hence $4$ hours are equal to $4$ half-lives. Therefore, the half-life of the substance is $1$ hour, which is an option (A). The decay constant $= \frac{\ln (2)}{h a l f l i f e} = \ln (2)$ per hour, which is an option (C).

Mean life $= \frac{1}{d e c a y c o n s \tan t}$

$= \frac{1}{\ln (2)}$ hour $= \frac{1}{0.693}$ hour

Hence, option (B) is incorrect.

After further $4$ hours (i.e., after $8$ hours), $\frac{1}{{(2)}^{8}} = \frac{1}{256} = \frac{100}{256} % = 0.39 %$ of the substance remains undecayed, which is an option (D). Thus the only incorrect option is (B).

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Physics Test - 27

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Physics Test - 27

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Question 1

5

4

3

2

Solution

Question 2

Solution

Question 3

50V

70V

130V

10V

Solution

Question 4

M

2Mπ

Mπ

Mπ

Solution

Question 5

4W1W2W1+W2

W1W2W1+W2

2W1W2W1+W2

W1+W22

Solution

Question 6

\(2\) seconds

\(3\) seconds

\(4\) seconds

\(5\) seconds

Solution

Question 7

H11

H2e4

F26e56

U23592

Solution

Question 8

F1F2

F2F1

F1F2F1+F2

F1+F2F1F2

Solution

Question 9

decreases continuously

decreases initially and increases again

remains unaltered

increases continuously

Solution

Question 10

The half-life of the sample is 1 hr.

The mean life of the sample is 12 hr.

The decay constant of the sample is (ln2)hr-1.

After further 4 hours, the amount of the substance leftover would be only 0.39% of the original amount.

Solution

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$5$

$4$

$3$

$2$

$50 V$

$70 V$

$130 V$

$10 V$

$M$

$\frac{2 M}{π}$

$\frac{M}{π}$

$M π$

$\frac{4 W_{1} W_{2}}{W_{1} + W_{2}}$

$\frac{W_{1} W_{2}}{W_{1} + W_{2}}$

$\frac{2 W_{1} W_{2}}{W_{1} + W_{2}}$

$\frac{W_{1} + W_{2}}{2}$

${}_{1}H^{1}$

${}_{2}H e^{4}$

${}_{26}F e^{56}$

${}_{92}U^{235}$

$\frac{F_{1}}{F_{2}}$

$\frac{F_{2}}{F_{1}}$

$\frac{F_{1} F_{2}}{F_{1} + F_{2}}$

$\frac{F_{1} + F_{2}}{F_{1} F_{2}}$

The half-life of the sample is $1$ hr.

The mean life of the sample is $\frac{1}{2}$ hr.

The decay constant of the sample is $(\ln 2) h r^{- 1}$ .

After further $4$ hours, the amount of the substance leftover would be only $0.39 %$ of the original amount.