Physics Test

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Physics Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

A bomber plane is moving horizontally with a speed of 600 m/s and a bomb released from it strikes the ground in 10 sec. What is the measure of angle at which the bomb strikes the ground? (g = 10 m/s²)

A

$\sin ^{-1}\left(\frac{1}{5}\right)$

B
tan^-1 (6)

C

$\tan ^{-1}\left(\frac{1}{5}\right)$

D

$\tan ^{-1}\left(\frac{1}{6}\right)$

Solution

Horizontal component of velocity will remain constant. $u_{x}=600 \mathrm{m} / \mathrm{s}$
Let h be the height of the plane from the ground. For the motion in vertical direction, $u_{y}=0$
then, $h=\frac{1}{2} g t^{2}$
$h=\frac{1}{2} \times 10 \times(10)^{2}=500 m$
Using $v_{y}^{2}-u_{y}^{2}=2 g h$
$v_{y}^{2}=2 \times 10 \times 500$
$v_{y}=100 \mathrm{m} / \mathrm{s}$
If the angle made by the resultant velocity with horizontal at the time of striking the ground is $\theta$.
$\tan \theta=\frac{v_{y}}{v_{x}}=\frac{100}{600}$
$\tan \theta=\frac{1}{6}$
Question 2
1 / -0

A block released on a rough inclined plane of inclination $\theta=30^{\circ}$ slides down the plane with an acceleration $g / 4$ where $g$ is the acceleration due to gravity. What is the coefficient of friction between the block and the inclined plane?

A
$\frac{2}{\sqrt{3}}$

B
$\frac{1}{\sqrt{3}}$

C

$\frac{1}{2 \sqrt{3}} $

D

$\frac{\sqrt{3}}{2}$

Solution

Acceleration of an object down the incline is 'a '

$\mathrm{ma}=\mathrm{mg} \sin \theta-\mathrm{f}$
$f=\mu N=\mu m g \cos \theta$
$a=g \sin \theta-\mu g \cos \theta$
$\theta=30^{\circ}$
$\frac{g}{4}=g\left(\sin 30^{\circ}-\mu \cos 30^{\circ}\right)$
$\frac{1}{2}-\mu \frac{\sqrt{3}}{2}=\frac{1}{4} \Rightarrow \mu \frac{\sqrt{3}}{2}=\frac{1}{4}$
$\mu=\frac{1}{2 \sqrt{3}}$
Question 3
1 / -0

A body is dropped from height 'h'. If the coefficient of restitution is 'e', calculate the height achieved (h₁) after one bounce.

A
h₁ = e²h

B
h₁ = e⁴h

C
h₁ = eh

D
h₁ = $\frac{h}{e}$

Solution

If a body falls from a height 'h' onto a hard floor and rebounds to height h₁, the coefficient of the restitution defined as the ratio of the velocity of approach to the velocity of separation.
Velocity of approach
$v = \sqrt{2 g h}$
Velocity of body after one bounce
$v_{1} = \sqrt{2 g h_{1}}$
Coefficient of restitution
$\begin{array}{l} e = \frac{v_{1}}{v} = \sqrt{\frac{2 g h_{1}}{2 g h}} \\ e = \sqrt{\frac{h_{1}}{h}} \end{array}$
Then,
$h_{1} = e^{2} h$
Question 4
1 / -0

To express the size of a molecule and an atom, a small size unit called the ____ is used.

A
parsec

B
light year

C
angstrom

D
nanometre

Solution

Angstromis a unit of length equal to 1/10,000,000,000 (one-ten billionth) of ametre (1 ^_$\times$10⁻¹⁰mor 100pm). Its symbol is theSwedish letter. $A^{0}$
The angstrom is often used in natural sciences and technology to express the size of atoms,molecules, microscopicbiological structures, the lengths ofchemical bonds, thearrangement of atoms in crystals, the wavelengths ofelectromagnetic radiation and the dimensions ofintegrated circuitparts.
Question 5
1 / -0

One femtometre is equivalent to

A
10^-15m

B
10¹⁵m

C
10^-12m

D
10¹²m

Solution

One femtometre is equivalent to 10^-15m.
Question 6
1 / -0

Which of the following units denotes the dimensions [ML²/Q²], where 'Q' represents the electric charge?

A
Wb/m²

B
Henry (H)

C
H/m²

D
Weber (Wb)

Solution

$\left[\mathrm{ML}^{2} \mathrm{Q}^{-2}\right]=\left[\mathrm{ML}^{2} \mathrm{A}^{-2} \mathrm{T}^{-2}\right]$
i)$[\mathrm{Wb}]=\left[\mathrm{ML}^{2} \mathrm{T}^{-2} \mathrm{A}^{-1}\right]$
ii)$\left[\mathrm{Wb} / \mathrm{m}^{2}\right]=\left[\mathrm{MT}^{-2} \mathrm{A}^{-1}\right]$
iii)[henry] $=\left[\mathrm{ML}^{2} \mathrm{T}^{-2} \mathrm{A}^{-2}\right]$
iv)$\left[\mathrm{H} / \mathrm{m}^{2}\right]=\left[\mathrm{M} T^{-2} \mathrm{A}^{-2}\right]$
So, henry (H) has dimensions $\mathrm{ML}^{2} / \mathrm{Q}^{2}$
Question 7
1 / -0

A solid sphere (M) attached to a massless spring (K) can roll without slipping along a horizontal surface. If the system is released after an initial stretch, the time period will be–

A
$2 π \sqrt{\frac{3 M}{2 K}}$

B
$2 π \sqrt{\frac{3 M}{5 K}}$

C
$2 π \sqrt{\frac{7 M}{5 K}}$

D
$2 π \sqrt{\frac{M}{K}}$

Solution
Question 8
1 / -0

A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of 10 cms^-1.

The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snapshot of the wave is shown in the figure. What is the velocity of point P when its displacement is 5 cm?

A

$-\frac{\sqrt{3} \pi}{50} \hat{\mathrm{j}} \mathrm{ms}^{-1}$

B
$\frac{\sqrt{3} \pi}{50} \hat{\mathrm{j}} \mathrm{ms}^{-1}$

C

$-\frac{\sqrt{3} \pi}{50} \hat{\mathrm{i}} \mathrm{ms}^{-1}$

D

$\frac{-\sqrt{3} \pi}{50} \hat{\mathrm{i}} \mathrm{ms}^{-1}$

Solution

$\mathrm{v}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{y}^{2}}$

$\mathrm{v}_{\mathrm{p}}=2 \pi \mathrm{f} \sqrt{\mathrm{A}^{2}-\mathrm{y}^{2}}$

$\mathrm{v}_{\mathrm{p}}=2 \pi \frac{\mathrm{c}}{\lambda} \sqrt{\mathrm{A}^{2}-\mathrm{y}^{2}}$

$\mathrm{v}_{\mathrm{p}}=\frac{2 \pi}{0.5} \times 0.1 \sqrt{(0.1)^{2}-(0.05)^{2}}$

= $\frac{\sqrt{3} \pi}{50} \hat{\mathrm{j}} \mathrm{ms}^{-1}$

Hence the correct answer is option (b).
Question 9
1 / -0

Four thin metal rods, each of mass M and length L, are welded to form a square ABCD as shown in the figure. What is the moment of inertia of the composite structure about a line which bisects rods AB and CD and is perpendicular to the plane of the structure?

A
$\frac{M L^{2}}{6}$

B
$\frac{M L^{2}}{3}$

C
$\frac{M L^{2}}{2}$

D
$\frac{2 M L^{2}}{3}$

Solution

Refer to Fig. Moment of inertia is a scalar quantity. So the moment of inertia of the structure is the sum of the moments of inertia of the four rods about the specified axis of rotation, i.e.,
$I=I_{1}+I_{2}+I_{3}+I_{4}$
where $I_{1}=$ moment of inertia of rod 1 about an axis passing through its centre $E$ and perpendicular to its plane $=\frac{M L^{2}}{12}$
$I_{2}=$ moment of inertia of rod 2 about an axis passing through its centre $F$ and perpendicular to its plane $=$ $\frac{M L^{2}}{12}$
$I_{3}=$ moment of inertia of rod 3 about a parallel axis at a distance $\frac{L}{2}$ from it $=M\left(\frac{L}{2}\right)^{2}=\frac{M L^{2}}{4},$ and
$I_{4}=$ moment of inertia of rod 4 about a parallel axis
at a distance $\frac{L}{2}$ from it $=\frac{M L^{2}}{4}$
$I =\frac{M L^{2}}{12}+\frac{M L^{2}}{12}+\frac{M L^{2}}{4}+\frac{M L^{2}}{4}$
$=\frac{2}{3} M L^{2}$

Hence the correct answer is option (d).
Question 10
1 / -0

A uniform cylindrical rod of mass 'M' and length 'L' is rotating with an angular speed 'ω'. The axis of rotation is perpendicular to its axis of symmetry and passes through one of its end faces. If the room temperature increases by 't' and the coefficient of linear expansion of the rod is 'α', the magnitude of change in its angular speed is

A
2 $α$ ωt

B
ωat

C
$\frac{3}{2}$ ωat

D
$\frac{w α t}{2}$

Solution

since, the moment of inertia of thin rod about an axis passing through its centre of mass and perpendicular to its geometrical axis is
$I_{C M}=\frac{M L^{2}}{12}$
M.I. of rod about an axis perpendicular to its geometrical axis and passes through one of the ends
$I_{1}=I_{C M}+M(L / 2)^{2}$
Theorem of parallel axis
$I_{1}=\frac{M L^{2}}{12}+\frac{M L^{2}}{4}$
$I_{1}=\frac{M L^{2}}{3}$

If initial angular speed of $\operatorname{rod}=\omega$ When room temperature increases by $t$ and coefficient of linear expansion of rod is $\alpha$ then New M. I. of rod
$I_{2}=\frac{M}{3}(L+l)^{2}$
where $l=L \times \alpha \times t$ (increment in length) No external torque is acting so angular speed will decrease. Let now angular speed $=(\omega-\Delta \omega)=\omega_{2}(\text { let })$ According to conservation of angular momentum (since, torque $\tau=0$ )
$I_{1} \omega_{1}=I_{2} \omega_{2}$
$\frac{M L^{2}}{3} \omega=\frac{M}{3}(L+l)^{2} \cdot(\omega-\Delta \omega)$
$L^{2} \omega=(L+L \cdot \alpha \cdot t)^{2}(\omega-\Delta \omega)$
$L^{2} \omega=L^{2}(1+\alpha \cdot t)^{2}(\omega-\Delta \omega)$
$\quad \frac{\omega-\Delta \omega}{\omega}=\frac{1}{(1+\alpha t)^{2}}$
$1-\frac{\Delta \omega}{\omega}=(1+\alpha t)^{-2}$
Using binomial theorem $(1+x)^{-n}$
$=1-n x+\ldots$
(leaving higher power terms) $1-\frac{\Delta \omega}{\omega}=1-2 \alpha t$
$\frac{\Delta \omega}{\omega}=2 \alpha t$
$\Delta \omega=2 \alpha t \omega$
$\therefore$ Change in angular speed $=2 \alpha t \omega$

Hence the correct answer is option (a).

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Physics Test - 5

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Physics Test - 5

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Question 1

\(\sin ^{-1}\left(\frac{1}{5}\right)\)

tan-1 (6)

\(\tan ^{-1}\left(\frac{1}{5}\right)\)

\(\tan ^{-1}\left(\frac{1}{6}\right)\)

Solution

Question 2

23

13

\(\frac{1}{2 \sqrt{3}} \)

\(\frac{\sqrt{3}}{2}\)

Solution

Question 3

h1 = e2h

h1 = e4h

h1 = eh

h1 =he

Solution

Question 4

parsec

light year

angstrom

nanometre

Solution

Question 5

10-15 m

1015 m

10-12 m

1012 m

Solution

Question 6

Wb/m2

Henry (H)

H/m2

Weber (Wb)

Solution

Question 7

2π3M2K

2π3M5K

2π7M5K

2πMK

Solution

Question 8

\(-\frac{\sqrt{3} \pi}{50} \hat{\mathrm{j}} \mathrm{ms}^{-1}\)

\(\frac{\sqrt{3} \pi}{50} \hat{\mathrm{j}} \mathrm{ms}^{-1}\)

\(-\frac{\sqrt{3} \pi}{50} \hat{\mathrm{i}} \mathrm{ms}^{-1}\)

\(\frac{-\sqrt{3} \pi}{50} \hat{\mathrm{i}} \mathrm{ms}^{-1}\)

Solution

Question 9

ML26

ML23

ML22

2ML23

Solution

Question 10

2αωt

ωat

32ωat

wαt2

Solution

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tan^-1 (6)

$\frac{2}{\sqrt{3}}$

$\frac{1}{\sqrt{3}}$

h₁ = e²h

h₁ = e⁴h

h₁ = eh

h₁ = $\frac{h}{e}$

10^-15m

10¹⁵m

10^-12m

10¹²m

Wb/m²

H/m²

$2 π \sqrt{\frac{3 M}{2 K}}$

$2 π \sqrt{\frac{3 M}{5 K}}$

$2 π \sqrt{\frac{7 M}{5 K}}$

$2 π \sqrt{\frac{M}{K}}$

$\frac{M L^{2}}{6}$

$\frac{M L^{2}}{3}$

$\frac{M L^{2}}{2}$

$\frac{2 M L^{2}}{3}$

2 $α$ ωt

$\frac{3}{2}$ ωat

$\frac{w α t}{2}$