Physics Test

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Physics Test - 6

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Weekly Quiz Competition

Question 1
3 / -1

A hollow spherical conductor of radius R is given a charge Q. Work done in moving a charge q from its centre to surface is :

A
Qq/4πε₀R

B
Qq/2πε₀R

C
Qq/πε₀R

D
zero

Solution

The electric field inside the hollow conducting sphere is zero. So there is no force inside the sphere. Thus the work done is zero for moving a charge q from center to surface of the sphere.

Hence option D is correct
Question 2
3 / -1

A flat car of mass M₀ starts moving to the right due to a constant horizontal force F at t = 0. Sand spills on the flat car from a stationary hopper. The velocity of loading is constant and is equal to $μ k g s^{- 1}$ Then,

A
kinetic energy of loaded car, at an instant, is less than the work done by the force f (up to that instant).

B
kinetic energy of loaded car at an instant is equal to work done by force F up to that instant.

C
Momentum of loaded car at an instant is less than impulse of force F up to that instant.

D
None

Solution

We see that the only force doing work in the direction of motion that is along X is the force. And from the work energy theorem, total work done is equal to change in kinetic energy.

Hence. KE of loaded car at any instant is equal to work done by force F till that instant.

Hence option B is correct.
Question 3
3 / -1

The ratio of the electrostatic force of attraction to the gravitational force between the proton and electron of the hydrogen atom is of the order of

A
10³⁹

B
10^-39

C
10⁸

D
10^-8

Solution

Electrostatic force of attraction $=\left( kq _{1} Q _{2}\right) / r _{2}$
$=\left[9 \times 10^{9} \times\left(1 \cdot 6 \times 10^{-19}\right)^{2}\right] / r^{2}$
Gravitational force of attraction $=\left( Gm _{ e } m _{ p }\right) / r ^{2}$
$=\left[\left(6 \cdot 67 \times 10^{-11}\right)\left(9 \cdot 1 \times 10^{-31}\right)\left(1 \cdot 67 \times 10^{-27}\right)\right] / r^{2}$
Ratiof $_{ e } / F _{ g }=\left[9 \times 10^{9} \times\left(1 \cdot 6 \times 10^{-19}\right)^{2}\right] /\left(6 \cdot 67 \times 10^{-11}\right)\left(9 \cdot 1 \times 10^{-31}\right)$
$\left(1 \cdot 67 \times 10^{-27}\right)$
$=\left(23 \cdot 04 \times 10^{-29}\right) /\left(101 \cdot 36 \times 10^{-69}\right)$
$=2 \cdot 27 \times 10^{39}$
Hence option A is correct.
Question 4
3 / -1

A proton of mass m charge e is released from rest in a uniform electric field of strength E. The time taken by it to travel a distance d in the field is :

A
√(2de/mE)

B
√(2dm/Ee)

C
√(2dE/me)

D
√(2Ee/dm)

Solution

force acting on proton =Eq.
acceleration a=Eq/m
given proton initialy at rest u=0
we know X=ut+(1/2)at2
⇒d=1/2⋅(Eq/m)t2
⇒t=√(2md/Eq)
⇒t=√(2md/Ee) ( q = e for a proton)
Hence option B is correct.
Question 5
3 / -1

Directions: Mark the correct statements.

A
Line spectra contains information about atoms only.

B
Line spectra contains information about both atoms and molecules.

C
Band spectra contains information about both atoms and molecules.

D
None of these

Solution

Line mission spectra can be obtained for both atoms and molecules, an atom or molecule in an excited state emits photons by making transition from excited state to ground state. Thus, constituting line emission spectra.

The wavelength emitted by molecular energy levels, which are generally grouped into several bunches, are also grouped and each group is well separated from each other.

The spectrum in this case looks like a band spectrum.

Hence option B is correct.
Question 6
3 / -1

Two alpha particles are separated by a distance of 10⁻¹³m. The force between them in free space is :

A
92⋅16×10⁻³N

B
92⋅16×10⁻⁴N

C
92⋅16×10⁻⁵N

D
92⋅16×10⁻²N

Solution

a particles are $He ^{+2}$ ions
$\therefore$ Net charge present on them is $2 \times\left(1 \cdot 6 \times 10^{-19}\right) C$.
We know $F =\left( kq _{1} q _{2}\right) / r _{2}$
$F =\left[\left(9 \times 10^{9}\right)\left(2 \times 1 \cdot 6 \times 10^{-19}\right)\left(2 \times 1 \cdot 6 \times 10^{-19}\right)\right] /\left(10^{-13}\right)^{2}$
$F =\left(92 \cdot 16 \times 10^{-29}\right) / 10^{-26}$
$F =92 \cdot 16 \times 10^{-3} N$

Hence option A is correct.
Question 7
3 / -1

Figure shows an electric dipole with positive charge at (0,b) and negative charge at (0,-b).
The electric potential magnitude along the x axis, from x=−a to x=a is correctly given by which graph?

A

B

C

D

Solution

As the charges are equidistant from x axis so the distance (r) at any point P on x-axis from two charges will be equal.
Thus, total potential at point P on x-axis is V=kQ/r+k(−Q)/r=0
Thus, option A will be correct.
Question 8
3 / -1

A stepped cylinder, with thread wound around smaller diametre, is released from rest and the cylinder moves down. Then,

A
frictional force, if any, acts up the include

B
coefficient of friction should be less than $\frac{\tan θ}{2}$

C
the cylinder cannot roll without slipping

D
All of the above

Solution

Acceleration of the point of contact, $a_{c}=a_{c n}+\alpha R$ Velocity of point of contact is non-zero. Hence, rolling with slipping will occur and friction will act upward. For translational motion, $m g \sin \theta-T-\mu m g \cos \theta=m a-\ldots-1$
For rotational motion, torque about CM of cylinder $-\mu m g \cos \theta \cdot R+T r=m K^{2} \alpha$
$\alpha=\frac{a}{r}[A s$ there is no slip of rope $]$ $-\mu m g \cos \theta \cdot R+T r=m K^{2} \frac{a}{r}$
$\mu m g \cos \theta \frac{R}{r}+T=m \frac{K^{2}}{r^{2}} a-\ldots-11$
From I and $\|, a=\frac{\sin \theta-m \cos \theta\left(1+\frac{B}{r}\right)}{1+\frac{K^{2}}{r^{2}}}$
$a>0 \sin \theta>\mu \cos \theta\left(1+\frac{R}{r}\right)$
$\mu<\frac{\tan \theta}{\left(1+\frac{\pi}{r}\right)}$
$A s R>r$
$\mu<\tan \frac{\theta}{2}$

Hence option D is correct.
Question 9
3 / -1

Figure shows lines of force for a system of two point charges. The possible choice for the charges is:

A
q₁=4 μC,q₂=−1.0 μC

B
q₁=1 μC,q₂=−4 μC

C
q₁=−2 μC,q₂=+4 μC

D
q₁=3 μC,q₂=2 μC

Solution

As lines of forces are coming out of q₁ and going into q₂

q₁→ positive charge

q₂→ negative charge

As number of lines of forces are greater near q₁, then:

∣q₁∣>∣q₂∣

Hence option A is correct.
Question 10
3 / -1

When a body of mass M is attached to lower end of wire (of length L) whose upper end is fixed, then the elongation of wire is l. Which of the following statements regarding this is/are correct?

A
Loss in gravitational potential energy of M is Mgl.

B
Elastic potential energy stored in wire is (Mgl/2).

C
Elastic potential energy stored in wire is Mgl.

D
A and B are correct

Solution

Loss in gratiational potential energy of M is Mgl, as M falls down by l.

Elastic potential energy =(1/2) x Stress x Strain x Volume

= (1/2) (Mg/A)x(I/L)x L

=(1/2) Mgl

Hence option d is correct.
Question 11
3 / -1

Positive point charges q₁ and q₂ are moving with velocities v₁ and v₂ as shown in the given figure. Mark the correct statements.

A
The magnetic field intensity at location of q₁ due to q₂ is zero

B
The magnetic field intensity at the location of q₂ due to q₁ is $\frac{μ_{0} q_{1} v_{1}}{4 π a}$

C
The magnetic force experienced by q₂ is q₂v₂ x103.JPG}} and is towards left

D
All of the above

Solution

Using Biot Savart law, magnetic field at location of $q _{2}$ due a $_{1}$
$B=\frac{\mu_{1} g_{1} v_{1}}{4 x_{2}^{2}} \sin 90^{\circ}=\frac{\mu_{1} \rho_{1} v_{1}}{4 \pi a^{2}},$ directed into plane of paper
Force experienced by q $_{2}$ due to magnetic field of $q_{1}$
$F = q _{2} V _{2}$
$B = q _{2} v _{2} \times \frac{ p _{0}}{4 \pi} \frac{q_{1} v_{1}}{a^{2}}$
The direction of force can be calculated using left hand rule, hence direction of force is towards left.
The charge $q_{1}$ is placed on line along which $q_{2}$ moves. Hence, magnetic field at location as qı due to $q _{2}$ is zero.
Thus, magnetic force experienced by 9 , is zero.

Hence option d is correct.
Question 12
3 / -1

Two identical pendulums A and B are suspended from the same point. Both are given a positive charge, with A having more charge than B. They diverge and reach equilibrium with the strings of pendulums A and B making angles θ1 and θ2 with the vertical respectively. Then:

A
θ₁>θ₂

B
θ₁<θ₂

C
θ₁=θ₂

D
the tension in A is greater than in B

Solution

Coulomb's electrostatic force acting on 2 charged particles is equal if they are kept isolated.
i.e it doesn't matter which bob has more charge, the force acting between them will still be the same and thus they will make the same angle from the vertical.
From Coulomb's law, the force acting on both the particles will be
F =(kQq)/r²
Hence , T₁=T₂ , which implies θ₁=θ₂
Hence option C is correct.
Question 13
3 / -1

A person connects four 1/4Ω cells in series but one cell has its terminal reversed. The external resistance is 1 Ω. If each cell has an emfof 1.5 V, the current flowing is:

A
(4/3)A

B
(3/4)A

C
1.5 A

D
zero

Solution

Given : r=1/4Ω E=1.5 volts R=1Ω
Equivalent potential difference of the circuit Eeq=E+E+E−E=2E=2×1.5=3 volts
Total resistance of the circuit
R_eq=4r+R=4×(1/4)+1=2Ω
∴ Current flowing through the circuit
I=E_eq/R_eq=3/2=1.5 A
Hence option C is correct.
Question 14
3 / -1

Two wires of same metal have the same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10Ω. The total resistance of the combination is:

A
(5/2)Ω

B
(40/3)Ω

C
40Ω

D
100Ω

Solution

Resistance of a wire R=ρl/A where ρ= resistivity, l= length, A= cross section of the wire.

As both have same material and length so R∝1/A

Thus, R₂/R₁=A1/A₂=3/1⇒R₂=3R₁.

here R₁ is the resistance of thicker wire so its resistance R₁=10Ω (given)

so, R₂=3(10)=30Ω

As they are connected in series so the equivalent resistance is

R_eq=R₁+R₂=10+30=40Ω

Hence option C is correct.
Question 15
3 / -1

An electric dipole of dipole moment 'p' is kept parallel to an electric field of intensity 'E'. The work done in rotating the dipole through an angle of 90° is :

A
Zero

B
-pE

C
pE

D
2pE

Solution

Work done in rotating a electric dipole of dipole moment p through angle θ from zero degree say is:
W=−pE[cosθ−cos0]=pE−pEcosθ
Here, θ is 90 therefore work done, W=pE
Hence option C is correct.