24 + Interview Questions in GENERAL QA IN NSEP Page 1 InterviewSolution

1.	In and atom an electron excites to the fourth orbit. When it jumps back to the energy levels a spectrum is formed. Total number of spectral lines in this spectrum would be (A) 3(B) 4(C) 5 (D) 6
Answer» Since there is only one atom, so number of line will be 3

Discussion

2.	A vertical narrow wire is illuminated with laser. Alternate dark and bright bands are formed on a graph paper pasted on a distant wall. Indicate the correct statement/s. (a) making appropriate measurements, it is possible to determine the diameter of the wire (b) this phenomenon exhibits that light does not follow rectilinear paths (c) this is a case of Fraunhoffer diffraction (d) this is a case of interference of an infinitely large number of Huygens's secondary waves leading to a diffraction pattern
Answer» Correct option (a)(b)(c)(d) Explanation: The observation indicated in the question are due to diffraction of light. When an obstacle of dimensions comparable to wavelength of light comes in the path of light, it bends around the obstacle and travels forward. During this process, the secondary wavelets interfere with each other to form a pattern of bright (constructive) and dark (destructive) bands on a screen. (a) Condition for the formation of dark band : a sin θ = mλ (here a is the dimensions of the obstacle : diameter of the thread in this case) (b) Diffraction is a wave phenomenon. Diffraction pattern is observed on the screen when light bends around an obstacle. (c) Diffraction can be studies in two ways : (1) Fraunhoffer diffraction : The obstacle, the source and the screen are far away from each other. Plane wave front approximation can be applied (2) Fresnel diffraction : The obstacle, the source and the screen are close to each other. Shape of wave front depends on the type of source. (spherical wave front for point source and cylindrical wave fronts for extended source) (d) Diffraction phenomenon inherently involves interference. Secondary wavelets from different waves interfere with each other to form alternate bright and dark bands of varying intensity on the screen.

2.

A vertical narrow wire is illuminated with laser. Alternate dark and bright bands are formed on a graph paper pasted on a distant wall. Indicate the correct statement/s. (a) making appropriate measurements, it is possible to determine the diameter of the wire (b) this phenomenon exhibits that light does not follow rectilinear paths (c) this is a case of Fraunhoffer diffraction (d) this is a case of interference of an infinitely large number of Huygens's secondary waves leading to a diffraction pattern

Answer»

Correct option (a)(b)(c)(d)

Explanation:

The observation indicated in the question are due to diffraction of light. When an obstacle of dimensions comparable to wavelength of light comes in the path of light, it bends around the obstacle and travels forward.

During this process, the secondary wavelets interfere with each other to form a pattern of bright (constructive) and dark (destructive) bands on a screen.

(a) Condition for the formation of dark band : a sin θ = mλ (here a is the dimensions of the obstacle : diameter of the thread in this case)

(b) Diffraction is a wave phenomenon. Diffraction pattern is observed on the screen when light bends around an obstacle.

(c) Diffraction can be studies in two ways :

(1) Fraunhoffer diffraction : The obstacle, the source and the screen are far away from each other. Plane wave front approximation can be applied

(2) Fresnel diffraction : The obstacle, the source and the screen are close to each other. Shape of wave front depends on the type of source. (spherical wave front for point source and cylindrical wave fronts for extended source)

(d) Diffraction phenomenon inherently involves interference. Secondary wavelets from different waves interfere with each other to form alternate bright and dark bands of varying intensity on the screen.

Discussion

3.

Which one of the following statements is not a tautology ?(1) P → (p ∨ q)(2) (p ∧ q) → (~p) ∨ q(3) (p ∧ q) → p (4) (p ∨ q) → (p) ∨ (~q)

Answer»

Correct option (4) (p ∨ q) → (p) ∨ (~q)

Explanation:

P	q	p V q	∨ q	p ∨ ~q	(P ∨ q) → (p) ∨ (~q)
T	T	T	F	T	T
T	F	T	T	T	T
F	T	T	F	F	F
F	F	F	T	T	T

Discussion

4.	Which one of the following devices does not respond to the intensity of light incident on it ?(A) Photoresistor (LDR)(B) Photodiode(C) Light Emitting Diode(D) Solar Cell
Answer» (c) Theory based

Discussion

5.	The equation correctly represented by the following graph is (a and b are constants)(A) x + y = b(B) ax2 + by2 = 0 (C) x + y = ab (D) y = axb
Answer» Correct option (D) y = ax^b Explanation: log y = m log x + C log y = logc' x^m y = c'x^m y = ax^b

5.

The equation correctly represented by the following graph is (a and b are constants)(A) x + y = b(B) ax2 + by2 = 0 (C) x + y = ab (D) y = axb

Answer»

Correct option (D) y = ax^b

Explanation:

log y = m log x + C

log y = logc' x^m

y = c'x^m

y = ax^b

Discussion

6.	A ball is dropped onto a horizontal surface from a height of 36 cm. After bouncing several times it comes to rest covering a total distance of 100 cm measured in a vertical direction. The percentage loss in its kinetic energy after its first impact is(a) 36 (b) 64 (c) 53 (d) 96
Answer» Correct option (c) 53 Explanation: One can write the series 100 = 36 + 2 x 36 (e² + e⁴ + e⁶ + ..) giving e² = 8/17 where e is the coefficient of restitution. Now, the percentage loss of kinetic energy = change in kinetic energy/original kinetic energy x 100. Note that the speed after impact is e times the initial speed. This gives the percentage loss as 52.94 which is almost 53 %.

6.

A ball is dropped onto a horizontal surface from a height of 36 cm. After bouncing several times it comes to rest covering a total distance of 100 cm measured in a vertical direction. The percentage loss in its kinetic energy after its first impact is(a) 36 (b) 64 (c) 53 (d) 96

Answer»

Correct option (c) 53

Explanation:

One can write the series 100 = 36 + 2 x 36 (e² + e⁴ + e⁶ + ..) giving e² = 8/17 where e is the coefficient of restitution. Now, the percentage loss of kinetic energy = change in kinetic energy/original kinetic energy x 100. Note that the speed after impact is e times the initial speed. This gives the percentage loss as 52.94 which is almost 53 %.

Discussion

7.	The frequency of the pendulum having maximum amplitude is (a) f0 / 2 (b) f0 (c) 2 f0 (d) not related to f0
Answer» Correct option (b) f₀ Explanation: Frequency of the driver pendulum : f₀ (given) The driven pendulum having maximum amplitude is in resonance with the driving pendulum. Resonance is achieved when the natural frequency of driven pendulum is equal to the frequency of the driving pendulum. So the frequency of the pendulum having maximum amplitude is f₀

7.

The frequency of the pendulum having maximum amplitude is (a) f0 / 2 (b) f0 (c) 2 f0 (d) not related to f0

Answer»

Correct option (b) f₀

Explanation:

Frequency of the driver pendulum : f₀ (given)

The driven pendulum having maximum amplitude is in resonance with the driving pendulum. Resonance is achieved when the natural frequency of driven pendulum is equal to the frequency of the driving pendulum.

So the frequency of the pendulum having maximum amplitude is f₀

Discussion

8.	Which of the following will give a straight line graph? (a) log A against t (b) log A against1/t (c) A against t2 (d) A2 against t
Answer» Correct option (a) Explanation: Consider the damped oscillation : A = A₀ e-bt/2m Apply ln on both sides : ln A = ln A₀ - b/2m t The equation is in the form of y = - mx + c This is a straight line with negative slope and positive y intercept. So the correct option is (a)

8.

Which of the following will give a straight line graph? (a) log A against t (b) log A against1/t (c) A against t2 (d) A2 against t

Answer»

Correct option (a)

Explanation:

Consider the damped oscillation :

A = A₀ e-bt/2m

Apply ln on both sides : ln A = ln A₀ - b/2m t

The equation is in the form of y = - mx + c

This is a straight line with negative slope and positive y intercept. So the correct option is (a)

Discussion

9.	The variation of amplitude A with respect to time t is shown as
Answer» Correct option (b) Explanation: Ideal case: no damping: the amplitude remains constant. (third graph) Practical case: damping is present: amplitude decreases exponentially. Assuming the damping force is directly proportional to velocity of the bob. We can write the amplitude : A = A₀ e- bt/2m Here b is damping constant and m is mass of the pendulum bob. So the correct option is (b)

9.

The variation of amplitude A with respect to time t is shown as

Answer»

Correct option (b)

Explanation:

Ideal case: no damping: the amplitude remains constant. (third graph)

Practical case: damping is present: amplitude decreases exponentially. Assuming the damping force is directly proportional to velocity of the bob. We can write the amplitude : A = A₀ e- bt/2m

Here b is damping constant and m is mass of the pendulum bob.

So the correct option is (b)

Discussion

10.	An object is placed 30 cm away from a symmetric convex lens and an image two thirds of the size of the object is produced. The object is moved by a distance of 20 cm so as to get a magnified image. Now we get (a) a real image of magnification (b) a virtual image of magnification 5 (c) a real image at a distance of 40 cm (d) a virtual image at a distance of 60 cm
Answer» Correct option (d) a virtual image at a distance of 60 cm Explanation: Size of the image : size of the object ⇒ v = 2/3 u Lens formula (real image) : 1/v - 1/u = 1/f ⇒ 3/2u + 1/u = 1/f ⇒ f = 2u/5 = 2 x 30 /5 = 12 cm After moving the object by 20 cm, it is with in the focus of the lens. So it forms a virtual image on the same side as that of the object. Lens formula (virtual image : 1/v - 1/u = 1/f ⇒ - 1/v + 1/10 = 1/12 ⇒ 1/v = 1/60 ⇒ v = 60 cm

10.

An object is placed 30 cm away from a symmetric convex lens and an image two thirds of the size of the object is produced. The object is moved by a distance of 20 cm so as to get a magnified image. Now we get (a) a real image of magnification (b) a virtual image of magnification 5 (c) a real image at a distance of 40 cm (d) a virtual image at a distance of 60 cm

Answer»

Correct option (d) a virtual image at a distance of 60 cm

Explanation:

Size of the image : size of the object ⇒ v = 2/3 u

Lens formula (real image) : 1/v - 1/u = 1/f

⇒ 3/2u + 1/u = 1/f

⇒ f = 2u/5 = 2 x 30 /5 = 12 cm

After moving the object by 20 cm, it is with in the focus of the lens. So it forms a virtual image on the same side as that of the object.

Lens formula (virtual image : 1/v - 1/u = 1/f

⇒ - 1/v + 1/10 = 1/12

⇒ 1/v = 1/60

⇒ v = 60 cm

Discussion

11.	Consider two point masses m1 and m2 connected by alight rigid rod of length r0. The moment of inertia of the system about an axis passing through their centre of mass and perpendicular to the rigid rod is given by
Answer» Correct option (B) ur₀²

11.

Consider two point masses m1 and m2 connected by alight rigid rod of length r0. The moment of inertia of the system about an axis passing through their centre of mass and perpendicular to the rigid rod is given by

Answer»

Correct option (B)

ur₀²

Discussion

12.	A particle moves in a plane with a constant speed along a path y = 2x2 + 3x – 4. When the particle is at (0, – 4) the direction along which it is moving is inclined to the X axis at an angle (a) 630 (b) 720 (c) 270 (d) 00
Answer» Correct option (b) 72⁰ Explanation: The direction along which the particle moves is the direction of velocity which in turn, is the slope of the trajectory. This is [dy/dx]_(0,4) = tan(θ) = 3 and hence the answer.

12.

A particle moves in a plane with a constant speed along a path y = 2x2 + 3x – 4. When the particle is at (0, – 4) the direction along which it is moving is inclined to the X axis at an angle (a) 630 (b) 720 (c) 270 (d) 00

Answer»

Correct option (b) 72⁰

Explanation:

The direction along which the particle moves is the direction of velocity which in turn, is the slope of the trajectory. This is [dy/dx]_(0,4) = tan(θ) = 3 and hence the answer.

Discussion

13.	Let a body be placed at a point on the earth’s surface at a latitude λ where the radius of the earth is R. Then, the body experiences an effective acceleration(a) g - Rω2 cos λ(b) g + Rω2 sin λ(c) g - Rω2 cos2 λ(d) g - Rω2/ cos2λ
Answer» Correct option (c) g - Rω² cos² λ Explanation: Effective acceleration g' is the vector sum of g and acceleration due to rotation of the earth rω² where r is the distance from the axis of rotation. Also one can neglect rω² compared to g while determining the magnitude g'. Therefore, g' = (g² - 2grω²cos² λ)^1/2. Also r can be written as Rcosλ where λ is the latitude.

13.

Let a body be placed at a point on the earth’s surface at a latitude λ where the radius of the earth is R. Then, the body experiences an effective acceleration(a) g - Rω2 cos λ(b) g + Rω2 sin λ(c) g - Rω2 cos2 λ(d) g - Rω2/ cos2λ

Answer»

Correct option (c) g - Rω² cos² λ

Explanation:

Effective acceleration g' is the vector sum of g and acceleration due to rotation of the earth rω² where r is the distance from the axis of rotation. Also one can neglect rω² compared to g while determining the magnitude g'. Therefore, g' = (g² - 2grω²cos² λ)^1/2. Also r can be written as Rcosλ where λ is the latitude.

Discussion

14.	Two new and different cells having emf v1 and v2 have their balance points P1 and P2 respectively, then(a) v1 > v2 (b) v1 < v2 (c) v1 = v2 (d) insufficient data
Answer» Correct option (b) v₁ < v₂ Explanation: Principle of potentiometer: emf of the cell connected in secondary circuit of the potentiometer is directly proportional to the balancing length of the potentiometer. The balancing length is measured from that point where the positive terminal of the standard cell is connected. From the diagram, we can see that point P₁ is closer to the starting point than the point P₂

14.

Two new and different cells having emf v1 and v2 have their balance points P1 and P2 respectively, then(a) v1 > v2 (b) v1 < v2 (c) v1 = v2 (d) insufficient data

Answer»

Correct option (b) v₁ < v₂

Explanation:

Principle of potentiometer:

emf of the cell connected in secondary circuit of the potentiometer is directly proportional to the balancing length of the potentiometer.

The balancing length is measured from that point where the positive terminal of the standard cell is connected.

From the diagram, we can see that point P₁ is closer to the starting point than the point P₂

Discussion

15.	The best material for strips marked S and S1 is(a) Plastic (b) Aluminium (c) Cast Iron (d) Plated Brass
Answer» Correct option (d) Plated Bras Explanation: Brass is the best conducting material out of the available options. As it offers minimum resistance, the readings will be close to actual values.

15.

The best material for strips marked S and S1 is(a) Plastic (b) Aluminium (c) Cast Iron (d) Plated Brass

Answer»

Correct option (d) Plated Bras

Explanation:

Brass is the best conducting material out of the available options. As it offers minimum resistance, the readings will be close to actual values.

Discussion

16.	It is required to decide the ratio (e1 / e2) by sum and difference method also. Four cells with different values of emf E are available. One must use a cell of emf E equal to (a) 1.40 V (b) 2.0 V (c) 4.5 V (d) 1.08 V
Answer» Correct option (c) 4.5 V Explanation: Ratio of emfs : e₁/e₂ = l₁/l₂ Ratio of emfs by sum and difference method : e_{1 +}e₂/e₁- e₂ = l₁/l₂ = e₁/e₂ = l₁+ l₂ /l₁- l₂ The emf of the cell used in the primary circuit should be greater than (e₁ + e₂) E > (1.4 + 1.08) E > 2.48 volt

16.

It is required to decide the ratio (e1 / e2) by sum and difference method also. Four cells with different values of emf E are available. One must use a cell of emf E equal to (a) 1.40 V (b) 2.0 V (c) 4.5 V (d) 1.08 V

Answer»

Correct option (c) 4.5 V

Explanation:

Ratio of emfs : e₁/e₂ = l₁/l₂

Ratio of emfs by sum and difference method : e_{1 +}e₂/e₁- e₂ = l₁/l₂ = e₁/e₂ = l₁+ l₂ /l₁- l₂

The emf of the cell used in the primary circuit should be greater than (e₁ + e₂) E > (1.4 + 1.08)

E > 2.48 volt

Discussion

17.	It is claimed that the strips S and S1 serve two purposes I. to decide the end points of the wire and II. To keep the wires in place, then (a) both I and II are important (b) I is more important than II (c) II is more important than I (d) both are equally important
Answer» Correct option (b) I is more important than II Explanation: When the end points of the wire are known accurately, the readings of the potentiometer will be trustworthy. So it is more important to decide the end points of the wire than to keep the wires in place.

17.

It is claimed that the strips S and S1 serve two purposes I. to decide the end points of the wire and II. To keep the wires in place, then (a) both I and II are important (b) I is more important than II (c) II is more important than I (d) both are equally important

Answer»

Correct option (b) I is more important than II

Explanation:

When the end points of the wire are known accurately, the readings of the potentiometer will be trustworthy. So it is more important to decide the end points of the wire than to keep the wires in place.

Discussion

18.	A convex lens and concave lens are kept in contact and the combination is used for the formation of image of a body by keeping it at different places on the principal axis. The image formed by this combination of lenses can be : (A) Magnified, inverted and real(B) Diminished, inverted and real(C) Diminished, erect and virtual(D) Magnified, erect and virtual
Answer» Correct option (A)(B)(C)(D) Explanation: Combination may behave converging In that case (a), (b) & (d) are possible . If combination behaves like diverging c will be correct. So all the options are correct.

18.

A convex lens and concave lens are kept in contact and the combination is used for the formation of image of a body by keeping it at different places on the principal axis. The image formed by this combination of lenses can be : (A) Magnified, inverted and real(B) Diminished, inverted and real(C) Diminished, erect and virtual(D) Magnified, erect and virtual

Answer»

Correct option (A)(B)(C)(D)

Explanation:

Combination may behave converging In that case (a), (b) & (d) are possible . If combination behaves like diverging c will be correct. So all the options are correct.

Discussion

19.	Refer to the circuit given below. Output voltage V0 is measured between points a and b. Then,(a) the peak value of V0 is 2.5 V above the minimum if the diode is assumed to be ideal (b) the positive half cycle of the input is clipped (c) the circuit acts as a rectifier (d) the peak value V0 is about 3.2 V above the minimum if D is silicon diode (non-ideal)
Answer» Correct option (a) the peak value of V₀ is 2.5 V above the minimum if the diode is assumed to be ideal (c) the circuit acts as a rectifier (d) the peak value V₀ is about 3.2 V above the minimum if D is silicon diode (non-ideal) Explanation: The diode is reverse biased in the +ve half cycle. The diode is forward biased in the –ve half cycle. (a) If the diode is ideal, if offers infinite resistance in reverse bias, thus the entire input appears across the output. (2.5 V above the minimum) (c) Since the negative half cycle is clipped, the circuit acts as rectifier. (d) For a non-ideal silicon diode, threshold voltage is 0.7 V, so the output voltage will be (0.7 + 2.5 = 3.2 V) above the minimum.

19.

Refer to the circuit given below. Output voltage V0 is measured between points a and b. Then,(a) the peak value of V0 is 2.5 V above the minimum if the diode is assumed to be ideal (b) the positive half cycle of the input is clipped (c) the circuit acts as a rectifier (d) the peak value V0 is about 3.2 V above the minimum if D is silicon diode (non-ideal)

Answer»

Correct option (a) the peak value of V₀ is 2.5 V above the minimum if the diode is assumed to be ideal (c) the circuit acts as a rectifier (d) the peak value V₀ is about 3.2 V above the minimum if D is silicon diode (non-ideal)

Explanation:

The diode is reverse biased in the +ve half cycle.

The diode is forward biased in the –ve half cycle.

(a) If the diode is ideal, if offers infinite resistance in reverse bias, thus the entire input appears across the output. (2.5 V above the minimum)

(c) Since the negative half cycle is clipped, the circuit acts as rectifier.

(d) For a non-ideal silicon diode, threshold voltage is 0.7 V, so the output voltage will be (0.7 + 2.5 = 3.2 V) above the minimum.

Discussion

20.	If the threshold of hearing is assumed to be the reference (0 dB), then the threshold of pain is taken to be 120 dB. Let the corresponding sound intensities be I0 and I respectively. Then I0/I is(a) 120 (b) 1012 (c) 10-12 (d) 101.2
Answer» Correct option (C) 10^-12 Explanation: Use the definition of dB scale: 120 = 10 log₁₀(I/I₀) , which gives (I₀/I) = 10^-12 .

20.

If the threshold of hearing is assumed to be the reference (0 dB), then the threshold of pain is taken to be 120 dB. Let the corresponding sound intensities be I0 and I respectively. Then I0/I is(a) 120 (b) 1012 (c) 10-12 (d) 101.2

Answer»

Correct option (C) 10^-12

Explanation:

Use the definition of dB scale: 120 = 10 log₁₀(I/I₀) , which gives (I₀/I) = 10^-12 .

Discussion

21.	An isolated metallic object is charged in vacuum to a potential V0 using a suitable source, its electrostatic energy being W0. It is then disconnected from the source and immersed in a large volume of dielectric with dielectric constant K. The electrostatic energy of the sphere in the dielectric is :(A) K2W0(B) K W0(C) W0/K2(D) W0/K
Answer» Correct option (D) W₀/K Explanation: U = Q² /2C since C will become k times So U will become 1/k times

21.

An isolated metallic object is charged in vacuum to a potential V0 using a suitable source, its electrostatic energy being W0. It is then disconnected from the source and immersed in a large volume of dielectric with dielectric constant K. The electrostatic energy of the sphere in the dielectric is :(A) K2W0(B) K W0(C) W0/K2(D) W0/K

Answer»

Correct option (D) W₀/K

Explanation:

U = Q² /2C since C will become k times

So U will become 1/k times

Discussion

22.	A large cylindrical vessel contains water to a height of 10 m. It is found that the thrust acting on the curved surface is equal to that at the bottom. If atmospheric pressure can support a water column of 10 m, the radius of the vessel is(a) 10 m (b) 15 m (c) 5 m (d) 25 m
Answer» Correct option (b) 15 m Explanation: Thrust on the curved surface 2πRh (P₀ + h/2) = 300πR here as the thrust on the bottom = πR²h(P₀ + h/2) = 20πR². The atmospheric pressure in terms of column of water is 10 m. equating the two thrusts gives R = 15 m

22.

A large cylindrical vessel contains water to a height of 10 m. It is found that the thrust acting on the curved surface is equal to that at the bottom. If atmospheric pressure can support a water column of 10 m, the radius of the vessel is(a) 10 m (b) 15 m (c) 5 m (d) 25 m

Answer»

Correct option (b) 15 m

Explanation:

Thrust on the curved surface 2πRh (P₀ + h/2) = 300πR here as the thrust on the bottom = πR²h(P₀ + h/2) = 20πR². The atmospheric pressure in terms of column of water is 10 m. equating the two thrusts gives R = 15 m

Discussion

23.	A thin annular metal disc of inner and outer radii R1 and R2 respectively, is freely suspended from a point on its outer circumference. The length of the corresponding equivalent simple pendulum is
Answer» Correct option (b) Explanation: Moment of inertia of annular disc about the axis under consideration can be written as m/2(R²₁ + R²₂) + mR²₂ Comparing the expression for the periodic time of this oscillating disc as a compound pendulum with that of a simple pendulum, we get the equivalent length.

23.

A thin annular metal disc of inner and outer radii R1 and R2 respectively, is freely suspended from a point on its outer circumference. The length of the corresponding equivalent simple pendulum is

Answer»

Correct option (b)

Explanation:

Moment of inertia of annular disc about the axis under consideration can be written as m/2(R²₁ + R²₂) + mR²₂ Comparing the expression for the periodic time of this oscillating disc as a compound pendulum with that of a simple pendulum, we get the equivalent length.

Discussion

24.	Two identical thin metal strips, one of aluminum and the other of iron are riveted together to form a bimetallic strip. The temperature is raised by 50°C. If the central planes of the two strips are separated by 2 mm and the coefficients of thermal expansion for aluminum and iron are respectively 30 x 10-6 / °C and 10 x 10-6 / °C, the average radius of curvature of the bimetallic strip is about (a) 50 cm (b) 100 cm (c) 150 cm (d) 200 cm
Answer» Correct option (d) 200 cm Explanation: For aluminum one can write L_Al = L₀[1 + α_Alt] = R₂θ where R₂ is the radius of the arc of aluminum strip and t is the temperature. Note that θ is the angular width of the strips. A similar expression can be written for iron strip. The difference between the radii, say d, for aluminum and iron is the distance between the central planes of the two strips. The expression for d can be obtained from the above two expressions as = L_0t[α_Al - α_Fe]/d .Also (R₁ + R₂) can be approximated as 2i and the value of R can be determined.

24.

Two identical thin metal strips, one of aluminum and the other of iron are riveted together to form a bimetallic strip. The temperature is raised by 50°C. If the central planes of the two strips are separated by 2 mm and the coefficients of thermal expansion for aluminum and iron are respectively 30 x 10-6 / °C and 10 x 10-6 / °C, the average radius of curvature of the bimetallic strip is about (a) 50 cm (b) 100 cm (c) 150 cm (d) 200 cm

Answer»

Correct option (d) 200 cm

Explanation:

For aluminum one can write L_Al = L₀[1 + α_Alt] = R₂θ where R₂ is the radius of the arc of aluminum strip and t is the temperature. Note that θ is the angular width of the strips. A similar expression can be written for iron strip. The difference between the radii, say d, for aluminum and iron is the distance between the central planes of the two strips. The expression for d can be obtained from the above two expressions as = L_0t[α_Al - α_Fe]/d .Also (R₁ + R₂) can be approximated as 2i and the value of R can be determined.

Discussion

Explore topic-wise InterviewSolutions in .

In and atom an electron excites to the fourth orbit. When it jumps back to the energy levels a spectrum is formed. Total number of spectral lines in this spectrum would be (A) 3(B) 4(C) 5 (D) 6

Which one of the following statements is not a tautology ?(1) P → (p ∨ q)(2) (p ∧ q) → (~p) ∨ q(3) (p ∧ q) → p (4) (p ∨ q) → (p) ∨ (~q)

Which one of the following devices does not respond to the intensity of light incident on it ?(A) Photoresistor (LDR)(B) Photodiode(C) Light Emitting Diode(D) Solar Cell

The equation correctly represented by the following graph is (a and b are constants)(A) x + y = b(B) ax2 + by2 = 0 (C) x + y = ab (D) y = axb

A ball is dropped onto a horizontal surface from a height of 36 cm. After bouncing several times it comes to rest covering a total distance of 100 cm measured in a vertical direction. The percentage loss in its kinetic energy after its first impact is(a) 36 (b) 64 (c) 53 (d) 96

The frequency of the pendulum having maximum amplitude is (a) f0 / 2 (b) f0 (c) 2 f0 (d) not related to f0

Which of the following will give a straight line graph? (a) log A against t (b) log A against1/t (c) A against t2 (d) A2 against t

The variation of amplitude A with respect to time t is shown as

Consider two point masses m1 and m2 connected by alight rigid rod of length r0. The moment of inertia of the system about an axis passing through their centre of mass and perpendicular to the rigid rod is given by

A particle moves in a plane with a constant speed along a path y = 2x2 + 3x – 4. When the particle is at (0, – 4) the direction along which it is moving is inclined to the X axis at an angle (a) 630 (b) 720 (c) 270 (d) 00

Let a body be placed at a point on the earth’s surface at a latitude λ where the radius of the earth is R. Then, the body experiences an effective acceleration(a) g - Rω2 cos λ(b) g + Rω2 sin λ(c) g - Rω2 cos2 λ(d) g - Rω2/ cos2λ

Two new and different cells having emf v1 and v2 have their balance points P1 and P2 respectively, then(a) v1 &gt; v2 (b) v1 &lt; v2 (c) v1 = v2 (d) insufficient data

The best material for strips marked S and S1 is(a) Plastic (b) Aluminium (c) Cast Iron (d) Plated Brass

It is required to decide the ratio (e1 / e2) by sum and difference method also. Four cells with different values of emf E are available. One must use a cell of emf E equal to (a) 1.40 V (b) 2.0 V (c) 4.5 V (d) 1.08 V

It is claimed that the strips S and S1 serve two purposes I. to decide the end points of the wire and II. To keep the wires in place, then (a) both I and II are important (b) I is more important than II (c) II is more important than I (d) both are equally important

If the threshold of hearing is assumed to be the reference (0 dB), then the threshold of pain is taken to be 120 dB. Let the corresponding sound intensities be I0 and I respectively. Then I0/I is(a) 120 (b) 1012 (c) 10-12 (d) 101.2

A large cylindrical vessel contains water to a height of 10 m. It is found that the thrust acting on the curved surface is equal to that at the bottom. If atmospheric pressure can support a water column of 10 m, the radius of the vessel is(a) 10 m (b) 15 m (c) 5 m (d) 25 m

A thin annular metal disc of inner and outer radii R1 and R2 respectively, is freely suspended from a point on its outer circumference. The length of the corresponding equivalent simple pendulum is

Two new and different cells having emf v1 and v2 have their balance points P1 and P2 respectively, then(a) v1 > v2 (b) v1 < v2 (c) v1 = v2 (d) insufficient data