Physics Test

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Physics Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Copper and iron wires of same length and diameter are in series and connected across a battery. The resistivity of copper is about one-sixth of the iron. If E₁ and E₂ are the electric fields in the copper and iron wires respectively, then which of the following is correct?

A
E₁< E₂

B

C
E₁>E₂

D
E₁=E₂=0 because electric field can not exist in metals

Solution
Question 2
1 / -0

On a hypothetical planet satellite can only revolve in quantized energy level i.e. magnitude of energy of a satellite is integer multiple of a fixed energy. If two successive orbit have radius R and $\frac{3 R}{2}$ what could be maximum radius of satellite

A
9R

B
6R

C
4R

D
3R

Solution

For energy in radius $r _{ i }=\left|\frac{- GMm }{2 R }\right|= nk$
Where $n$ is integer and $k$ is constant energy,
now
$\frac{ GMm }{2 R }= nk$
$\frac{ GMm }{2\left(\frac{3 R }{2}\right)}=( n -1) k$
So solving $k=\frac{G M m}{6 R}$
Now this implies that for $R _{ max }, K =1$ $\frac{ GMm }{2 R _{\max }}=(1) \frac{ GMm }{6 R } \Rightarrow R _{\max }=3 R$

Hence option D is the correct answer.
Question 3
1 / -0

Two bodies A and B have emissivity’s 0.5 and 0.8 respectively. At some temperatures, the two bodies have maximum spectral emissive powers at wavelength 8000 A^o and 4000 A^o respectively. The ratio of their emissive powers at these temperatures is:

A
5/128

B
10

C
5/16

D
None of these

Solution

Let the body have temperatures $T_{1}$ and $T_{2}$ respectively at wavelength $\lambda_{1}=8000 A$ and $\lambda_{2}=4000$ A. .: From Wien's displacement law $\lambda T = constant$
$\Rightarrow \lambda_{1} T _{1}=\lambda_{2} T _{2}$ or $8000 \times T _{1}=4000 T _{2}$
or $\quad \frac{ T _{1}}{ T _{2}}=\frac{1}{2}$
Emissive power $=$ e $\sigma AT ^{4}$
$\therefore$ Ratio of emissive powers at these temperatures is $\frac{e_{1} T_{1}^{4}}{e_{2} T_{2}^{4}}=\frac{0.5}{0.8} \times\left(\frac{1}{2}\right)^{4}=\frac{5}{128}$

Hence option A is the correct answer.
Question 4
1 / -0

In the circuit shown in figure find the current in branch AB of the circuit :

A
5 A

B
0.5 A

C
11/3 A

D
None of these

Solution

In the circuit equivalent, Resistance across the battery is $R_{e q}=\frac{40}{11} \Omega$

Thus current through the battery is $I=\frac{20}{40 / 11}=5.5 A$
Thus current $1.5 A$ (from the figure) will be divided in $10 \Omega \& 5 \Omega$ in inverse ratio thus $15 \Omega=\frac{1.5 \times 10}{15}=1 A$
Thus current in branch $AB$ is
$1_{A B}=1+4=5 A$

Hence option A is the correct answer.
Question 5
1 / -0

A point object moves on a circular path such that distance covered by it is given by function
$S = (\frac{t^{2}}{2} + 2 t)$ Meter (t in second). The ratio of the magnitude of acceleration at t = 2 sec.
And t = 5 sec. is 1: 2 then radius of the circle is

A
1m

B
$3 \sqrt{51} m$

C
$\sqrt{51} m$

D
3 m

Solution

\begin{equation}\begin{array}{l}
\begin{array}{ll}
a_{t}=1 m / s^{2} & v=t+2 \\
\text { at } t=2 \sec & a(2)=\sqrt{1^{2}+\left\{\frac{(2+2)^{2}}{R}\right\}} \\
& a(5)=\sqrt{1^{2}+\left\{\frac{(5+2)^{2}}{R}\right\}^{2}} \\
2 \sqrt{1+\frac{16^{2}}{R^{2}}}=\sqrt{1+\frac{49^{2}}{R^{2}}} \\
4+\frac{(4) 16^{2}}{R^{2}}=1+\frac{49^{2}}{R^{2}}
\end{array} \\
3=\frac{49^{2}}{R^{2}}-4 \frac{16^{2}}{R^{2}} \Rightarrow R=3 \sqrt{51} m
\end{array}\end{equation}

Hence option B is the correct answer.
Question 6
1 / -0

A uniform electric field E is present horizontally along the paper throughout the region and uniform magnetic field B0 is present horizontally (perpendicular to plane of paper in inward direction) right to the line AB. A charge particle having charge q and mass m is projected vertically upward and it crosses the line AB after time t₀. Find the speed of projection if particle moves with constant velocity after t₀. (Given qE = mg)

A
gt₀

B
2gt₀

C
$\frac{g t_{0}}{2}$

D
Particle can’t move with constant velocity after crossing AB

Solution

Let velocity of particle at 'P' is 'v' making an angle ' $\theta$ ' with x-axis (P is the point where particle is crossing AB line) $\Rightarrow$ As after reaching point 'P', velocity of particle remain constant, so net force acting on particle will be zero at ${ }^{\prime} P ^{\prime}$
$So , F _{ B } \cos \theta= mg$
$\Rightarrow qv B _{0} \cos \theta= mg \ldots( i )$
and $F_{B} \sin \theta=F_{E}$
$\Rightarrow$ qv $B _{0} \sin \theta= qE$
$\Rightarrow E=v \cdot B_{0} \sin \theta \ldots(i i)$
Form (i) and (ii) $\tan \theta=\frac{ qE }{ mg }=\frac{ mg }{ mg }=1( qE = mg given )$
$\theta=45^{\circ}$
Now, horizontal velocity at point ${ }^{\prime} P ^{\prime}$ is due to Electric force q $E$ Only $So , V \cos \theta=\frac{ q E }{ m } t _{0}$
$=\frac{m g}{m} t_{0}$
$v \cos \theta=g t_{0} \ldots(i i i)$
vertical velocity at point ${ }^{\prime} P ^{\prime}$ $v \sin \theta=u-g t_{0} \ldots($ iv $)$
from (iii) and (iv)
\begin{equation}\begin{array}{l}
\tan \theta=\frac{u-g t_{0}}{g t_{0}} \\
\Rightarrow 1=\frac{u-g t_{0}}{g t_{0}} \\
\Rightarrow \quad u=2 g t_{0}
\end{array}\end{equation}

Hence option B is the correct answer.
Question 7
1 / -0

Consider a L-R circuit shown in figure. There is no current in circuit. Switch S is closed at t = 0, time instant when current in inductor is equal to current in resistor 2R will be :

A
$\frac{L}{R} \ln 2$

B
$\frac{2 L}{R} \ln 2$

C
$\frac{L}{2 R} \ln 2$

D
$\frac{L}{2 R}$

Solution
Question 8
1 / -0

A particle undergoes from position $O(0,0,0)$ to $A(a, 2 a, 0)$ via path $y=\frac{2 x^{2}}{a}$ in $x-y$ plane under the action of a force which varies with particle's $( x , y , z )$ coordinate as $\overrightarrow{ F }= x ^{2} yi + yz \hat{ j }+ xyz \hat{ k }$. Work done by the force $\overrightarrow{ F }$ is: (all symbols have their usual meaning and they are in SI unit.)

A
$\frac{4 a^{4}}{5}$

B
$\frac{a^{4}}{5}$

C
$\frac{a^{3}}{5}$

D
$\frac{2 a^{4}}{5}$

Solution

\begin{equation}
\begin{aligned}
&\vec{F}=x^{2} y \hat{i}+y z \hat{j}+x y z \hat{k}\\
&d r=d x \hat{i}+d y \hat{j}+d z k\\
&d w=F \cdot d \vec{r}\\
&d w=x^{2} y d x+y z d y+x y z d z\\
&\text { For given path, } z=0 \text { and } y=\frac{2 x^{2}}{a}\\
&\text { so, } d w=x^{2}\left(\frac{2 x^{2}}{a}\right) d x+0+0\\
&d w=\frac{2 x^{4}}{a} d x\\
&\boldsymbol{w}=\int \boldsymbol{d} w\\
&w=\int_{0}^{a} \frac{2 x^{4}}{a} d x\\
&\boldsymbol{w}=\frac{\overline{2}}{5 a}\left[\boldsymbol{x}^{5}\right]_{0}^{\boldsymbol{a}}\\
&w=\frac{2 a^{4}}{5}
\end{aligned}
\end{equation}

Hence option D is the correct answer.
Question 9
1 / -0

At time $t=0,$ a 2 kg particle has position vector $\vec{r}=(4 i-2 j)$ m relative to the origin. Its velocity is given by $\overrightarrow{ v }=2 t ^{2} \hat{ i }( m / s )$ for $t \geq 0 .$ The torque acting on the particle about the origin at $t =2 s ,$ is :

A
$32 N - m (\hat{k})$

B
$20 N - m (\hat{k})$

C
$16 N - m (\hat{k})$

D
$12 N - m (\hat{k})$

Solution
Question 10
1 / -0

An unknown particle $\frac{1285}{513} Z a$ originally at rest emits 5 alpha particles with speed 11385 km/h. Find the recoil speed of the unknown daughter nucleus.

A
35 m/s

B
50 m/s

C
134 m/s

D
100 m/s

Solution

when particle Za emits 5 alpha then
${ }_{513}^{1285} Za \rightarrow{ }_{503}^{1265} X+5\left({ }_{2}^{4} He \right)$
Using conservation of linear momentum
$1265 v^{\prime}+20 v=0$
$v^{\prime}=-\frac{20 v}{1265}$
$v^{\prime}=\frac{20 \times 11385}{1265}=180 km / h$
$v ^{\prime}=\frac{180 \times 1000}{60 \times 60}=50 m / s$

Hence option B is the correct answer.