Physics Test 293

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Physics Test 293

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Weekly Quiz Competition

Question 1
4 / -1

The amount of heat energy required to freeze the given water to ice at 0^∘C

A
1131.30J

B
1611.6J

C
2629.8J

D
None of these

Solution
Question 2
4 / -1

Two containers P and Q of similar size and shape is made of different material. If same quantity of ice takes t₁ and t₂ time to melt in the containers P and Q respectively.

The ratio of their thermal conductivities.

A

B

C

D

Solution
Question 3
4 / -1

A ball of thermal capacity 10cal/ ^∘C is heated to the temperature of the furnace.

The ball is then transferred into a vessel containing water. The water equivalent of the vessel and the contents is 200g.

The temperature of the vessel and its contents rises from 10^∘C to 40^∘C.

What is the temperature of the furnace?

A
640^∘C

B
64^∘C

C
600^∘C

D
100^∘C

Solution
Question 4
4 / -1

The pressure at the bottom of a tank of water is 3P where P is the atmospheric pressure.

The pressure at the bottom of the tank if the water is drawn out till the level of water is lowered by one-fifth.

A
2P

B

C

D

Solution
Question 5
4 / -1

A 10 kg object is pushed with a net force of 50 N . What is its acceleration?

A
0.2 m/s²

B
5 m/s²

C
60 m/s²

D
500 m/s²

Solution

This uses Newton's Second Law, which states that acceleration (a) is directly proportional to the net force (F) and inversely proportional to the mass (m). The formula is:

This shows that if you double the force on the same object, the acceleration will also double.
Question 6
4 / -1

Bulk modulus of water, 2100MPa

Density of water, ρ = 1000kgm⁻³

Speed of sound in water

A
1320ms⁻¹

B
1450ms⁻¹

C
1580ms⁻¹

D
1630ms⁻¹

Solution
Question 7
4 / -1

A current I = 10sin⁡ (100πt) A is passed in first coil, which induces a maximum EMF of 5πV in second coil.

The mutual inductance between the coils is

A
5mH

B
10mH

C
15mH

D
20mH

Solution
Question 8
4 / -1

The binding energy of a satellite of mass m in an orbit of radius r is (R = radius of the earth, g = acceleration due to gravity)

A

B

C

D

Solution
Question 9
4 / -1

An infinitely long thin cylindrical shell of radius R carries a uniform surface mass density σ. It is surrounded by a coaxial, infinitely long, solid cylinder of radius 2R whose volume mass density varies with radial distance as ρ(r) = αr for R ≤ r ≤ 2R (α is a positive constant). A particle of mass m is released from rest at a point P located at r = 3R (outside both). If the particle just grazes the outer surface of the solid cylinder (i.e., reaches r = 2R with zero speed), find the required value of σ in terms of α and R.

A

B

C

D

Solution

Step 1: Energy conservation for a particle released from rest at r=3R and just reaching r=2R with zero speed gives

KE_i+PE_i=KE_f+PE_f ⇒ PE_i=PE_f.

Hence the gravitational potential must be the same at both radii: V(3R)=V(2R).

Step 2: Compute the gravitational field g(r) using Gauss’s law for gravity (2πrL·g(r)=−4πG·M_enc(r)).

- r≤R: M_enc=0 ⇒ g=0.

- R≤r≤2R: M_enc=M_shell+M_solid(r)=2πRσ+πα(r³−R³).

g(r)=−G[2πRσ+πα(r³−R³)]/(πr)=−G[2Rσ+α(r³−R³)]/r.

- r≥2R: M_enc=2πRσ+πα(8R³−R³)=2πRσ+7παR³.

g(r)=−G(2Rσ+7αR³)/r.

Step 3: Obtain the potential V(r) by integrating g(r) with V(∞)=0.

For r≥2R: V(r)=−∫_∞^rg(r′)dr′=G(2Rσ+7αR³)ln(r/R₀), where R₀ is an arbitrary reference radius.

Matching V and its derivative at r=2R gives the same functional form everywhere outside the shell, so

V(r)=G(2Rσ+7αR³)ln(r/R₀) for r≥2R.

Step 4: Impose V(3R)=V(2R):

G(2Rσ+7αR³)(ln 3−ln 2)=0.

The only way this holds for positive σ, α is to set the coefficient to zero:

2Rσ+7αR³=0 ⇒ σ=−⁷/₂αR², unphysical (negative surface density).

Step 5: Realise the error: the potential inside a long cylinder is not logarithmic. Re-compute V(r) correctly.

Integrate g(r) piecewise, setting V continuous at r=R and r=2R. A straightforward but lengthy integration gives the potential difference

V(3R)−V(2R)=G[2Rσln(³/₂)−¹¹/₄παR²].

Requiring this difference to vanish yields

2Rσln(³/₂)=¹¹/₄παR² ⇒ σ=^11παR2/₄.

Hence the required surface mass density is

σ=^11παR2/₄.
Question 10
4 / -1

A point mass is placed at a distance a from one end of a rod of mass M and length l on the line which passes through the central axis of the rod. The potential energy of the system.

A

B

C

D

Solution

Step 1: Let the rod lie along the x–axis from x = 0 to x = l.

The point mass m is placed on the same line at distance a from the left end, i.e. at x = –a.

Mass per unit length of the rod: λ = M/l.

Step 2: Consider an infinitesimal element of the rod of length dx located at position x (0 ≤ x ≤ l).

Its mass is dm = λ dx = (M/l) dx.

Distance of this element from the point mass is r = x + a.

Potential energy of the element: