Physics Test 194

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Physics Test 194

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Question 1
4 / -1

A transverse wave is travelling along a horizontal string. The first picture shows the shape of the string at an instant of time. This picture is superimposed on a coordinate system to help you make any necessary measurements. The second picture is a graph of the vertical displacement of one point along the string as a function of time. How far does this wave travel along the string in one second?

A
0.3 cm

B
3.0 cm

C
9.0 cm

D
27 cm

Solution

From the graphs λ = 9cm

T = 3 sec
Question 2
4 / -1

A cyclic process of an enclosed gas of constant mass is represented by volume (V) against absolute temperature (T) as shown. If P represents pressure, the graph representing the same process can be

A

B

C

D

Solution

Co mb ina ti on of is ob ori c, is oc hor ic & isothermal.
Question 3
4 / -1

A closed organ pipe is vibrating in its second overtone. The length of the pipe is 10cm and maximum amplitude of vibration of particles of the air in the pipe is 2mm. Then the amplitude of S.H.M. of the particles at 9cm from the open end is:

A
√3 mm

B
√2 mm

C
√3/2 mm

D
none of these

Solution

4L/5 = λ ⇒ λ = 8cm

hus 2 cm corresponds to Δϕ = z/2

1 cm corresponds to Δϕ = z/4
Question 4
4 / -1

A sound source S and observers O₁, O₂ are placed as shown. S is always at rest and O₁, O₂ start moving with velocity v₀ at t = 0. At any later instant, let f₁ and f₂ represent apparent frequencies of sound received by O₁ and O₂, respectively. The ratio f₁/f₂ is

A
zero

B
between 0 and 1

C
1

D
> 1

Solution
Question 5
4 / -1

Equal masses of three liquids A, B and C have temperatures 10^oC, 25^oC and 40^oC respectively. If A and B are mixed, the mixture has a temperature of 15^oC. If B and C are mixed, the mixture has a temperature of 30^oC,. If A and C are mixed the mixture will have a temperature of

A
16^oC

B
20^oC

C
25^oC

D
29^oC

Solution

ms_A (15 – 10) = ms_B (25 – 15) s_A = 2s_B

ms_B (30 – 25) = ms_C (40 – 30)

s_B = 2s_C ⇒ s_A = 4s_C

ms_A (T – 10) = ms_C (40 – T)

⇒ 4(T – 10) = 40 – T

T = 16°C
Question 6
4 / -1

A steel rod is 4.000 cm in diameter at 30ºC. A brass ring has an interior diameter of 3.992 cm at 30ºC. In order that the ring just slides onto the steel rod, the common temperature of the two should be nearly (α_steel= 11 × 10^-6/ºC and α_brass= 19 × 10^-6/ºC)

A
200ºC

B
250ºC

C
280ºC

D
400ºC

Solution

For ring just slides on to the steel rod the diameter of rod and ring should be equal to each other and suppose due to Δθ increment in temperature the diameter of both are equal then
4 (1+ α_s Δθ) = 3.992 (1 + α_Brass Δθ) 4 + 4 × 11 × 10^-6 × Δθ
= 3.992 + 3.992 × 20 × 10^–6 × Δθ
4 + 44 × 10^–6 Δθ = 3.992 + 79.84 × 10^–6
× Δθ
0.008 = 35.84 × 10^–6 Δθ

so if temperature increased by 223°C then ring will start to slide and this temperature will equal to
θ = 30° + Δθ = 30 + 253 = 283°C
θ = 283°C ≈ 280°C
Question 7
4 / -1

A hot liquid is kept in a big room. The logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. The plot will be very nearly.

A
a straight line

B
a circular arc

C
a parabola

D
exponential decay

Solution

From N.Law of collision ℓn (T - T₀) = -kt + ℓn (T_i - T₀) (y = -mx + x) equation of straight line.
Question 8
4 / -1

A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C.The temperature of the surroundings is 20°C.

A
10 min

B
15 min

C
20 min

D
5 min

Solution
Question 9
4 / -1

A sinusoidal wave (longitudinal or transverse) is propagating through a medium in the direction of -ve x-axis. The parameters of the waves are A, ω and k. The particle at x = λ/4 executes the motion y(t) = A sinωt. Possible equation of the wave is

A
y(x, t) = A sin[ωt - kx + (π/2)]

B
y(x, t) = A sin[ωt + kx + (π/2)]

C
y(x, t) = A sin[ωt - kx - (π/2)]

D
y(x, t) = A sin[ωt + kx - (π/2)]

Solution

Let equation of wave as it is moving along - ve x-axis is y = A sin(kx + ωt +α) But, y(λ/4, t) = A sinωt Comparing then kx + α = 0 ⇒ a =-π/2
Question 10
4 / -1

PV versus T graph of equal masses of H₂, He and CO₂ is shown in figure.

Choose the correct alternative

A
3 corresponds to H₂, 2 to He and 1 to CO₂

B
1 corresponds to He, 2 to H₂ and 3 to CO₂

C
1 corresponds to He, 3 to H₂ and 2 to CO₂

D
1 corresponds to CO₂, 2 to H₂ and 3 to He

Solution

PV/T = tanθ = nR

∴ slope α no. of moles