Physics Test

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Physics Test - 24

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Question 1
1 / -0

The mirror in the below case is

A
Convex

B
Concave

C
Plane

D
None of these

Solution

Parallel ray before incidance is parallel after reflection, Hence mirror is plane

Hence, the correct option is (C)
Question 2
1 / -0

The combination of 'NAND' gates shown here (in figure), is equivalent to

A
An OR gate and an AND gate respectively

B
An AND gate and a NOT gate respectively

C
An AND gate and an OR gate respectively

D
An OR gate and a NOT gate respectively

Solution

\(\mathrm{C}=\overline{\mathrm{A}} \overline{\mathrm{B}} \quad\) (if a NAND is given only 1 input it behaves as not gate, as it is \(\overline{\mathrm{AA}}=\overline{\mathrm{A}}\) (one input is divided in two and then passed through NAND)

i.e. OR gate for second circuit \(C=\) Not of (A NAND B)
i.e. AND only
Hence, the correct option is (A)
Question 3
1 / -0

Two thin wire rings each having radius R are placed at a distance d apart with their axes coinciding. The charges on the two rings are \(+Q\) and - Q. The potential difference between the center of the two rings is :

A
1

B

C
zero

D

Solution

The electric potential at a distance \(\times\) from charged ring is \(V=\frac{q}{4 \pi \epsilon_{0} \sqrt{x^{2}+R^{2}}}\)
Thus, the potential at the center of ring of charge \(+Q=\) potential due to itself \(+\) potential due to
other ring of charge \(-Q\)
or \(V_{1}=\frac{Q}{4 \pi \epsilon_{0} \sqrt{0^{2}+R^{2}}}+\frac{(-Q)}{4 \pi \epsilon_{0} \sqrt{d^{2}+R^{2}}}\)
also the potential at the center of ring of charge \(-Q=\) potential due to itself \(+\) potential due to
other ring of charge \(+Q\) or \(V_{2}=\frac{(-Q)}{4 \pi \epsilon_{0} \sqrt{0^{2}+R^{2}}}+\frac{Q}{4 \pi \epsilon_{0} \sqrt{d^{2}+R^{2}}}\)
Thus, potential difference the center of the two rings is \(V=V_{1}-V_{2}\)
or \(V=\frac{Q}{4 \pi \epsilon_{0} R}+\frac{(-Q)}{4 \pi \epsilon_{0} \sqrt{d^{2}+R^{2}}}-\frac{(-Q)}{4 \pi \epsilon_{0} R}-\frac{Q}{4 \pi \epsilon_{0} \sqrt{d^{2}+R^{2}}}=\frac{Q}{2 \pi \epsilon_{0}}\left[\frac{1}{R}-\frac{1}{\sqrt{d^{2}+R^{2}}}\right]\)
Hence, the correct option is (B)
Question 4
1 / -0

A 0.5 kg block slides from A on horizontal track with an initial speed of3 m s^–1 towards a weightless horizontal spring of length 1 m1 m and force constant 2 N m^–1. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, the total distance through which the block moves before it comes to rest completely is -

A
2.5 m

B
4.42 m

C
4.24 m

D
2.44 m

Solution

At \(\mathrm{D}, \mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}-\mu_{\mathrm{k}} \mathrm{mg}(\mathrm{BD})\)
\(=\frac{1}{2} \times 0.5 \times 3^{2}-0.2 \times 0.5 \times 10 \times 2.14\)
\(=2.25-2.14=0.11 \mathrm{J}\)
\(0.11=\frac{1}{2} \mathrm{kx}^{2}+\mu_{\mathrm{R}} \mathrm{mg} \mathrm{x} \Rightarrow \mathrm{x}^{2}+\mathrm{x}-0.11=0\)
\(\Rightarrow \mathrm{x}=0.1 \mathrm{m}\)
Total distance covered \(=2+2.14+0.1=4.24 \mathrm{m}\)
Hence, the correct option is (C)
Question 5
1 / -0

The voltage of clouds is 4×10⁶ V with respect to ground. In a lightning strike lasting 100 ms, a charge of 4 C is delivered to the ground. The average power of lightning strike is (assume complete discharges )

A
160 MW

B
80 MW

C
20 MW

D
500 KW

Solution

Energy stored between cloud and earth, assuming the capacitor model is
\(W=\frac{g V}{2}\)
\(\frac{4 \times 10^{-5} \times 4}{2}=8 \times 10^{6} J\)
The power of lightning strike is
\(P=\frac{W}{t}=\frac{8 \times 10^{6}}{0.1}=80 \times 10^{6} W\)
\(=80 MW\)
Hence, the correct option is (B)
Question 6
1 / -0

An airplane requires a speed of 108 kmph to take off, to achieve that plane runs on the runway of 100 m. Mass of the plane is 10400 kg and the coefficient of friction between the plane and the ground is 0.2. Assuming the plane accelerates uniformly, the minimum force required is (Takeg=9.81 m/s²)

A
2×10⁴N

B
2.43×10⁴N

C
6.72×10⁴N

D
8.86×10⁴N

Solution

Required take off speed is \(108 \mathrm{kmph}\) i.e \(30 \mathrm{m} / \mathrm{s}\) over a run of \(100 \mathrm{m}\). \(v^{2}-u^{2}=2 a s\)
\(v^{2}-0=2 a s\)
\(a=\frac{v^{2}}{2 s}\)
Force that should be developed by the engine is given by:
\(F-\mu m g=m a\)
\(F=0.2(10400)(9.81)+10400\left(\frac{30^{2}}{2(100)}\right)=6.72 \times 10^{4} N\)
Hence, the correct option is (C)
Question 7
1 / -0

A particle moves under the effect of a force F=Cx from x=0 to x=x₁. The work done in the process is (treat C as a constant):

A

B

C

D

Solution
Question 8
1 / -0

A body freely falls from a certain height on to the ground in a time tt. During the first one-third of the time interval, it gains a kinetic energy Δk₁ and during the last one-third of the interval, it gains a kinetic energy Δk₂. The ratio Δk₁: Δk₂ is:

A
1:1

B
1:3

C
1:4

D
1:5

Solution

\(S=\frac{1}{2} g t^{2}\)
For the first one third of time:
\(\Delta K_{1} \propto\left(\frac{t}{3}\right)^{2}\)
For the last one third of time:
\(\Delta K_{2} \propto\left(t^{2}-\left(\frac{2 t}{3}\right)^{2}\right)\)
\(\frac{\Delta K_{1}}{\Delta K_{2}}=\frac{\left(\frac{t}{3}\right)^{2}}{\left(t^{2}-\left(\frac{2 t}{3}\right)^{2}\right)}=\frac{1}{5}\)
Hence, the correct option is (D)
Question 9
1 / -0

A body takes n times as much time to slide down an identical 45⁰ rough incline as it takes to slide down a smooth 45⁰ incline. The coefficient of friction is

A

B

C

D

Solution

For a smooth surface the acceleration along the surface is, \(a_{1}=g / \sqrt{2}\) So, \(t_{1}=\sqrt{\frac{2 s}{a_{1}}}\)
For the rough surface, acceleration is \(a_{2}=(1-\mu) g / \sqrt{2}\) \(t_{2}=n t_{1}\)
From above equations, we get \(\Rightarrow \mu=1-\frac{1}{n^{2}}\)
Hence, the correct option is (A)
Question 10
1 / -0

A particle is projected up along a rough plane of inclination 45⁰with the horizontal. If the coefficient of friction is 0.5, the retardation is (g = acceleration due to gravity):

A
g/2

B
g/2√2

C
3g/2√2

D
g/√2

Solution

Equating forces along \(\mathrm{X}\) and \(\mathrm{Y}\) axes, where \(m g \sin \theta\) is the parallel component of weight and \(m g \cos \theta\) is the normal
component \(N=m g \cos \theta\)
Also, \(F=m g \sin \theta+\mu_{k} m g \cos \theta\)
\(\therefore a=g \sin \theta+\mu_{k} g \cos \theta\)
\(\therefore a=g\left(\sin 45^{\circ}+0.5 \cos 45^{\circ}\right)\)
\(\therefore a=\frac{3 g}{2 \sqrt{2}}\)
Hence, the correct option is (C)