Mathematics Test

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Mathematics Test - 45

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Weekly Quiz Competition

Question 1
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A balloon is moving upwards with a speed of 20 m/s. When it is at a height of 14 m from ground in front of a plane mirror in as shown in figure, a boy drops himself from the balloon. Find the time duration for which he will see the image of source S placed symmetrically before plane mirror during free fall.

A
1.7 s

B
1.9 s

C
2.6 s

D
2.1 s

Solution

$\begin{aligned} \frac{\mathrm{PQ}}{\mathrm{BP}} &=\frac{\mathrm{IH}}{\mathrm{BH}}=\frac{\mathrm{HS}}{\mathrm{BH}} \\ \therefore \quad \mathrm{FQ} &=(\mathrm{BF})\left(\frac{\mathrm{HS}}{\mathrm{BH}}\right) \\=&(5)\left(\frac{1.0}{0.5}\right) \\=& 10 \mathrm{m} \\ \mathrm{FC} &=2+10=12 \mathrm{m} \end{aligned}$

The boy has dropped himself at point $F$. So, his velocity is $20 \mathrm{m} / \mathrm{s}$ in upward direction. Let us first find the time to move from $F$ to topmost point and then from topmost point to point $C$. From

\[
\begin{aligned}
\mathrm{s} &=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2} \text { we have } \\
-12 &=(20 t)+\frac{1}{2}(-10) \mathrm{t}^{2}
\end{aligned}
\]
Solving this equation we get, $t_{1}=4.53 \mathrm{s}$ Velocity of boy at point $Q$.
$v=\sqrt{(20)^{2}-2 \times 10 \times 10}=14.14 \mathrm{m} / \mathrm{s} \quad\left(v^{2}=u^{2}-2 \mathrm{gh}\right)$
Time taken to move the boy from $Q$ to topmost point and then from topmost point to $Q$ will be
$t_{2}=\frac{2 v}{\mathrm{g}}=2.83 \mathrm{s}$
$\therefore$ Time required time is $: t=t_{1}-t_{2}=1.7 \mathrm{s}$
Hence, the correct option is (A)
Question 2
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The peak emission from a body at a certain temperature occurs at a wavelength of $9000 Å .$ On increasing its temperature the total radiation emitted is increased to 81 times. At the initial temperature when the peak radiation from the black body is incident on a metal surface it does not cause any photoemission from the surface. After the increase of temperature the peak radiation from the black body causes photoemission. To bring these photoelectrons to rest, potential equivalent to the excitation energy between the n = 2 to n = 3 Bohr levels of hydrogen atom is required. Find the work function of the metal. [h = 6.62 × 10^-34 J-s and c = 3 × 10⁸ m/s]

A
2.25 eV

B
4.47 eV

C
7.72 eV

D
9.65 eV

Solution

Let $T_{1}$ be the initial tamprature and $T_{2}$ be the increased tamprature of the black body.
According to Stefan's law:
$\left(\frac{T_{2}}{T_{2}}\right)=81=(3)^{4}$
$T_{2}=3 T_{1}$
Also $\lambda_{1} T_{1}=i_{2} T_{2}$

$\lambda_{2}=\frac{\lambda_{1 \times I_{1}}}{T_{2}}=\frac{9000 \times T_{1}}{3 T_{1}}=3000$ A

Now $\frac{h v}{\lambda_{2}}-W=e V_{0}$ or $\frac{h v}{\lambda_{2}}-W=13.6\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$
Solving we get $, W=2.25 \mathrm{eV}$

Hence optioon A is the correct answer.
Question 3
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Q charge is uniformly distributed over the curve surface of a right circular cone of semi-vertical angle θ and height h. The cone is uniformly rotated about its axis at angular velocity ω. Calculated associated magnetic dipole moment.(H>R)

A
$\frac{q \omega h^{2} \tan ^{2} \theta}{4}$

B
$\frac{\tan ^{2} \theta q \omega h^{2}}{2}$

C
$\frac{\tan ^{2} \theta q \omega h^{2}}{3}$

D
$\frac{\tan ^{2} \theta q \omega h^{2}}{1}$

Solution

$A=$ Surface area of cone
$=\pi R \sqrt{R^{2}+h^{2}}$
$d A=2 \pi y d x$

$\tan \theta=\frac{y}{x}$
$y=x \tan \theta$
$d q=\frac{q}{A} d A$
$d i=d q \frac{\omega}{2 \pi}$
$=\frac{q}{\pi R \sqrt{R^{2}+h^{2}}} \frac{\omega}{2 \pi} \times 2 \pi y d x$
$d m=d i A$
$=\frac{q \omega}{\pi R \sqrt{R^{2}+h^{2}}} x \tan \theta d x \times \pi y^{2}$
$=\frac{q \omega}{\pi R \sqrt{R^{2}+h^{2}}} x \tan \theta d x \times \pi \times x^{2} \tan ^{2} \theta$
$\int d m=\frac{q \omega}{R \sqrt{R^{2}+h^{2}}} \int_{x=0}^{x=h} x^{3} \tan ^{3} \theta d x$
$=\frac{q \omega}{R \sqrt{R^{2}+h^{2}}} \times \frac{h^{4}}{4} \tan ^{3} \theta$
$=\frac{q \omega h^{4} \tan ^{4} \theta}{4 R \sqrt{R^{2}+h^{2}}}$
$=\frac{q \omega h^{4} \tan ^{3} \theta}{4 R h}(\because h>>R)$
$=\frac{q \omega h^{2} \tan ^{2} \theta}{4}$
key concept
Magnetic dipole and dipole moment
Magnetic Dipole Moment
From the expression for the torque on a current loop, the characteristics of the current loop are summarized in its magnetic moment

If there are $\mathrm{N}$ loops, then $\mu=\mathrm{NIA}$
The magnetic moment can be considered to be a vector quantity with direction perpendicular to the current loop in the right-hand-rule direction.
Hence, the correct option is (A)
Question 4
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A beam of ultraviolet radiation having wavelength between 100 nm and 200 nm is incident on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state, which wavelengths will have low intensity in the transmitted beam ?

A
104 nm

B
103 nm

C
105 nm

D
100 nm

Solution
Question 5
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A large tank filled with water to a height 'h' is to be emptied through a small hole at the bottom. The ratio of times taken for the level of water to fall from $\mathrm{h}$ to $\frac{\mathrm{h}}{2}$ and from $\frac{\mathrm{h}}{2}$ to zero is :

A
$\sqrt{2}$

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

C
$\sqrt{2} - 1$

D
$\frac{1}{\sqrt{2} - 1}$

Solution

Time taken for the level to fall from $\mathrm{H}$ to $\mathrm{H}^{\prime}$
$\mathrm{t}=\frac{\mathrm{A}}{\mathrm{A}_{0}} \quad \sqrt{\frac{2}{\mathrm{g}}}\left[\sqrt{\mathrm{H}}-\sqrt{\mathrm{H}^{\prime}}\right]$
According to problem the time taken for the level to fall from h to $\frac{h}{2}$
$t_{1}=\frac{A}{A_{0}} \quad \sqrt{\frac{2}{g}}\left[\sqrt{h}-\sqrt{\frac{h}{2}}\right]$
and similarly time taken for the level to fall from $\frac{h}{2}$
$t_{2}=\frac{A}{A_{0}} \quad \sqrt{\frac{2}{g}}\left[\sqrt{\frac{h}{2}}-0\right]$
$\therefore \frac{t_{1}}{t_{2}}=\frac{1-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}-0}=\sqrt{2}-1$
Hence, the correct option is (C)
Question 6
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Initially car A is 10.5 m ahead of car B. Both start moving at time t = 0 in the same direction along a straight line. The velocity time graph of two cars is shown in figure. The time when the car B will catch the car A, will be :

A
t = 21 sec

B
t=2 $\sqrt{5}$ sec

C
t = 20 sec

D
t = 1 sec

Solution
Question 7
1 / -0

The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of maximum intensity to minimum intensity in the interference fringe pattern observed.

A
32

B
36

C
38

D
34

Solution

When the flat car collides with the bumper, due to inertia of motion the bob swings forward No work is done by tension of string on the bob therefore energy is conserved

$\mathrm{KE}_{\mathrm{A}}+\mathrm{PE}_{\mathrm{A}}=\mathrm{KE}_{\mathrm{B}}+\mathrm{PE}_{\mathrm{B}}$

$v_{0}^{2}=2 \mathrm{g} l(1-\cos \theta)$

$v_{0}^{2}=4 \mathrm{g} l \sin ^{2} \theta / 2$

$\therefore \theta=2 \sin ^{-1}\left(\frac{w_{0}}{2 \sqrt{g l}}\right)$

For,

$\theta=60^{\circ}, l=10 \mathrm{m}, \mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2}$

$v_{0}^{2}=4 \times 10 \times 10 \sin ^{2}(30)=100$

$v_{0}=10 \mathrm{m} / \mathrm{s}$
Hence, the correct option is (D)
Question 8
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A truck starts from origin, accelerating with 'a' m/sec² in positive x - axis direction. After 2 seconds a man standing at the starting point of the truck projected a ball at an angle 30⁰ with velocity 'v' m/s. Find the relation between 'a' and 'v' such that ball hits the truck. (assume truck is moving on horizontal plane and man projected the ball from the same horizontal level of truck).

A

$\nu=2 \mathrm{g}\left[\frac{\sqrt{\frac{\mathrm{a}}{\mathrm{B} \sqrt{3}}}}{1-\sqrt{\frac{\mathrm{s}}{\mathrm{g} \sqrt{3}}}}\right]$

B

$\nu=2 \mathrm{g}\left[\frac{\sqrt{\frac{\mathrm{a}}{\mathrm{g}}}}{1-\sqrt{\frac{\mathrm{a}}{\mathrm{g}}}}\right]$

C

$\nu=g\left[\frac{\sqrt{\frac{a}{g \sqrt{3}}}}{1-\sqrt{\frac{a}{g \sqrt{3}}}}\right]$

D
$\nu=g\left[\frac{\sqrt{\frac{a}{8}}}{1-\sqrt{\frac{a}{g}}}\right]$

Solution

$\nu \sin 30^{\circ}(t-2)-\frac{1}{2} g(t-2)^{2}=0$
$\nu \cos 30^{\circ}(t-2)=\frac{1}{2} a t^{2}$
$\tan 30^{\circ}=\frac{\mathrm{g}(\mathrm{t}-2)^{2}}{\mathrm{at}^{2}} \Rightarrow \frac{\mathrm{t}-2}{\mathrm{t}}=\sqrt{\frac{\mathrm{a}}{\mathrm{g} \sqrt{3}}}$
$\Rightarrow t\left(1-\sqrt{\frac{a}{g \sqrt{3}}}\right)=2$
$\Rightarrow t=\frac{2}{1-\sqrt{\frac{3}{8 \sqrt{3}}}}$
$\Rightarrow \nu=\frac{g(t-2)}{2 \sin 30^{\circ}}$
$\Rightarrow g\left[\frac{2}{1-\sqrt{\frac{2}{8 \sqrt{3}}}}-2\right]$
$\nu=2 \mathrm{g}\left[\frac{\sqrt{\frac{0}{8 \sqrt{3}}}}{1-\sqrt{\frac{2}{8 \sqrt{3}}}}\right]$
Hence, the correct option is (A)
Question 9
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A battery is supplying power to a tape-recorder by cable of resistance of 0.2 $Ω$ . If the battery is generating 50 W power at 5V, then power received by tape-recorder is :

A
50 W

B
45 W

C
30 W

D
48 W

Solution

P = VI, $50 = 5 \times I$
$I = 10 A$

Power lost in cable= I²R = 10 × 10 × 0·2 = 2W

Power supplied to T.R= 50 W - 2W = 48W

Concepts :
Main Concept :
Electric PowerThe rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.
$P=\frac{W}{t}=V \cdot i=i^{2} R=\frac{V^{2}}{R}$
Units: It's S.I. unit is Joule/sec or Watt
Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt
Hence, the correct option is (D)
Question 10
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Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are) correct ?

A
Both cylinders P and Q reach the ground at the same time.

B
Cylinders P has larger linear acceleration than cylinder Q.

C
Both cylinders reach the ground with same translational kinetic energy.

D
Cylinder Q reaches the ground with larger angular speed.

Solution

a=MgsinθM+1R₂

ap=MgsinθM+MR₂R₂≈g2sinθ

aQ=gsinθasIQ∼0

mgh=12Iω₂+12mv₂

v=Rω

Undefined control sequence \therefore

Undefined control sequence \therefore

Undefined control sequence \therefore

Concepts :
Main Concept :
Calculation of moment of inertiaSince the moment of inertia of an ordinary object involves a continuous distribution of mass at a continually varying distance from any rotation axis, the calculation of moments of inertia generally involves calculus, the discipline of mathematics which can handle such continuous variables. Since the moment of inertia of a point mass is defined by
$I = m r^{2}$
Then the moment of inertia contribution by an infinitesimal mass element dm has the same form. This kind of mass element is called a differential element of mass and its moment of inertia is given by

$d I = r^{2} d m$
The “d” preceding any quantity denotes a vanishingly small or “differential” amount of it
Note that the differential element of moment of inertia dI must always be defined with respect to a specific rotation axis. The sum over all these mass elements is called an integral over the mass.
$\mathrm{I}=\int \mathrm{dI}=\int_{0}^{\mathrm{M}} \mathrm{r}^{2} \mathrm{dm}$
Hence, the correct option is (D)