100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Two element M and N belong to group I and II respectively and are in the same period of the periodic table. How do the following properties of M and N vary?1) Size of their atoms2) This metallic characters3) Their valencies informing oxides.4) Molecular, formulae of their chlorides.
Answer» 1. M and N belong to the same period but group I and II. Therefore N will be smaller than M as a atomic size is decreases from left to right. 2. M is more metallic than N. Metallic character goes on decreasing from left to right as tendency to lose electrons. 3. Their valencies are 1 and 2 respectively in forming oxides. Valency goes on increasing first and then decreases. 4. MCl, NCl₂ are molecules formulae of their chlorides.

Discussion

2.	What type of dielectric material is used in capacitors used for fans and for power factor correction? (a) Oil impregnated paper (b) Vacuum (c) Glass (d) Mica
Answer» Correct option: (a) Oil impregnated paper

Discussion

3.	What creates washing action in washing machine? (a) energy (b) water supply (c) turbulence (d) Electricity
Answer» Correct option: (c) turbulence

Discussion

4.	If field current is decreased in shunt dc motor, the speed of the motor (a) Decrease (b) Increase (c) Remains same (d) None of the above
Answer» Correct option: (b) Increase

Discussion

5.	Which of the following motor has negative speed regulation? (a) Series motor (b) Stunt motor (c) Cumulative compound motor (d) Differential compound motor
Answer» Correct option: (d) Differential compound motor

Discussion

6.	The motor used in household refrigerators is (a) dc series motor (b) dc shunt motor (c) universal motor (d) single phase induction motor.
Answer» Correct option: (d) single phase induction motor.

Discussion

7.	Which motor cannot be started on no load? (a) Series Motor (b) Shunt motor (c) Cumulative compound motor (d) Both b and c.
Answer» Correct option: (a) Series Motor

Discussion

8.	A body of mass 2 kg moving with a velocity `3 m//sec` collides with a body of mass 1 kg moving with a velocity of `4 m//sec` in opposite direction. If the collision is head-on perfect inelastic, thenA. both particles will move together with velocity `(2//3) m//sec` after collisionB. the momentum of the system is `2 kg m//sec` throughoutC. the momentum of the system is `10 kg m//sec` throughoutD. The loss in KE of the system is `(49//3)` Joule
Answer» Correct Answer - A::B::D Applying momentum conservation. `m_(1)u_(1) - m_(2) u_(2) = (m_(1) + m_(2))v` `2 (3) - 1 (4) = (2 + 1) v` `rArr v = (2)/(3)` m/sec Net mometum `= m_(1) u_(1) - m_(2) u_(2)` `2 (3) - 1 (4) = 2 kg-m//sec` Loss in `KE = (1)/(2) m_(1) u_(1)^(2) + (1)/(2) m_(2) u_(2)^(2) - (1)/(2) (m_(1) + m_(2)) v^(2)` `= (1)/(2) xx 2 xx (3)^(2) + (1)/(2) xx (1) xx (4)^(2) - (1)/(2) (2 + 1) ((2)/(3))^(2)` `= 9 + 8 - (2)/(3) = (49)/(3)` Joule

Discussion

9.	In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas (a) the temperature will decrease, (b) the volume will increase (c) the pressure will remain constant (d) the temperature will increase
Answer» Correct answer is (a) From first law of thermodynamics dQ = dU + dW dQ = dU (if dW = 0) Since, dQ < 0 dU < 0 or U_final < U_initial Hence, temperature will decrease.

9.

In a given process of an ideal gas, dW = 0 and dQ < 0. Then for the gas (a) the temperature will decrease, (b) the volume will increase (c) the pressure will remain constant (d) the temperature will increase

Answer»

Correct answer is (a) From first law of thermodynamics

dQ = dU + dW

dQ = dU (if dW = 0)

Since, dQ < 0

dU < 0

or U_final < U_initial

Hence, temperature will decrease.

Discussion

10.	It’s no use shouting at grandmother, she’s as deaf as a ________. A) post B) pin C) pole D) door E) wall
Answer» Correct option is A) post

Discussion

11.	If you leave that butter in the sun, it will ________. A) thaw B) dissolve C) set D) harden E) melt
Answer» Correct option is E) melt

Discussion

12.	(i) \( H _{2}( g )+ Cl _{2}( g ) \longrightarrow 2 HCl \) (g); \( \Delta H =- x kJ \) (ii) \( NaCl + H _{2} SO _{4} \longrightarrow NaHSO _{4}+ HCl ; \Delta H =- y kJ \) (iii) \( 2 H _{2} O +2 Cl _{2} \longrightarrow 4 HCl + O _{2} ; \Delta H =- z kJ \) From the above equations, the value of \( \Delta H _{ f } \) of \( HCl \) is (1) \( -x kJ \) (2) \( - y kJ \) (3) \( - z k J \) (y) \( -x / 2 k J \)
Answer» As we know, the formation of one mole of a compound from its elements in their most stables states of aggregation is called molar enthalpy of formation. Thus, in equation (1) HCl is formed from most stable elemental state of Hydrogen & chlorine. H₂(g) + Cl₂(g) → 2HCl. \(\Delta\)H₁= -x KJ. ∴ formation of one mole of HCl- \(\frac12\)H₂(g) + \(\frac12\)Cl₂(g) → HCl \(\Delta H\) = \(\frac{\Delta H_1}2\) = \(-\frac{x}2\) KJ Hence, \(\Delta\)H_f of HCl will be -x/2

12.

(i) $ H _{2}( g )+ Cl _{2}( g ) \longrightarrow 2 HCl $ (g); $ \Delta H =- x kJ $ (ii) $ NaCl + H _{2} SO _{4} \longrightarrow NaHSO _{4}+ HCl ; \Delta H =- y kJ $ (iii) $ 2 H _{2} O +2 Cl _{2} \longrightarrow 4 HCl + O _{2} ; \Delta H =- z kJ $ From the above equations, the value of $ \Delta H _{ f } $ of $ HCl $ is (1) $ -x kJ $ (2) $ - y kJ $ (3) $ - z k J $ (y) $ -x / 2 k J $

Answer»

As we know, the formation of one mole of a compound from its elements in their most stables states of aggregation is called molar enthalpy of formation.

Thus, in equation (1) HCl is formed from most stable elemental state of Hydrogen & chlorine.

H₂(g) + Cl₂(g) → 2HCl. $\Delta$H₁= -x KJ.

∴ formation of one mole of HCl-

$\frac12$H₂(g) + $\frac12$Cl₂(g) → HCl $\Delta H$ = $\frac{\Delta H_1}2$ = $-\frac{x}2$ KJ

Hence, $\Delta$H_f of HCl will be -x/2

Discussion

13.	Select the incorrect option `:`A. Sodium peroxide dissolves in water giving `H_(2)O_(2)` and `NaOH`.B. Both `LiNO_(3)` and `NaNO_(3)` on heating separately decompose and each liberates two gases `NO_(2)` and `O_(2)^(-)`C. Solvay process can not be used for the manufacture of potassium hydrogen carbonate.D. Alkali metals are prepared only by the electrolysis of their fused chlorides.
Answer» Correct Answer - 2 `4LiNO_(3)overset(Delta)rarr 2Li_(2)O+4NO_(2)+O_(2)` `2NeNO_(3)overset(Delta)rarr2NaNO_(2)+O_(2)`

Discussion

14.	In FCC lattice of NaCl structure, if the diameter of `Na^(+)` is `x,` and the radius of `Cl^(-)` is y, then the edge length of the crystal is `:`A. `2x+2y`B. `x+2y`C. `(x)/(2)+y`D. None of the above.
Answer» Correct Answer - B In FCC lattice of NaCl structure `Cl^(-)` are place at the corner of edge length and `Na^(+)` is place at octahedral void `i.e.` middle of two corner `Cl^(-)` ion.

Discussion

15.

What is the difference between (Helium-neon)he-ne laser and ruby laser?

Answer»

The difference between Helium-Neon Laser and Ruby laser are given below:

Helium-Neon Laser	Ruby Laser
These are gaseous State laser.	They are solid-state laser.
It requires electrical pumping.	It requires optical pumping.
The laser output is continuous in it.	It is in the form of pulse in the ruby laser.
It has high efficiency.	It has low efficiency.
There are no any defects due to crystalline imperfections.	The defects due to crystalline imperfections are present in it.

Discussion

16.	Which resonating structure is most stable?A. B. C. D.
Answer» Dut to less distance between two oppostiting charge and its aromatic nature and `NO_(2)^(-)` alos have-1 effect which stabilize it.

Discussion

17.	Statement-l: In Ellingham diagram metal oxide, having higher value of AG is more stable, than metal oxide having less AG.Statement II → In Ellingham diagram lower metal will reduce uppers metal from it's oxide select true (T) and false (F) regarding these statements respectively. (1) T, T(2) F, F(3) T, F(4) F. T
Answer» Correct option is (4) F, T Higher value of ΔG like HgO, Ag₂O are less stable. In Ellingham diagram lower situated metals is more reactive, it can reduce higher metal oxide.

17.

Statement-l: In Ellingham diagram metal oxide, having higher value of AG is more stable, than metal oxide having less AG.Statement II → In Ellingham diagram lower metal will reduce uppers metal from it's oxide select true (T) and false (F) regarding these statements respectively. (1) T, T(2) F, F(3) T, F(4) F. T

Answer»

Correct option is (4) F, T

Higher value of ΔG like HgO, Ag₂O are less stable.

In Ellingham diagram lower situated metals is more reactive, it can reduce higher metal oxide.

Discussion

18.	If 4 litre of `H_(2)` gas at 400 mm Hg and `47^(@)C` is transferred to 19 litre flask at `107^(@)C`. Then pressure of `H_(2)` gas is:A. 191.7mm of HgB. 100 mm of HgC. 158.4 mm of HgD. 200 mm of Hg
Answer» `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `(400xx4)/(320)=(P_(2)xx19)/(180)rArrP_(2)=100mm` of Hg

Discussion

19.	Which of the following statement is/are false ?A. `Delta_rS` for `1/2Cl_2(g)toCl(g)` is positiveB. `DeltaElt0` for combustion of `CH_4(g)` in a sealed container with rigid adiabatic systemC. `DeltaG` is always zero for a reversible process in a closed systemD. `DeltaG^@` for an ideal gas reaction is a function of pressure
Answer» Correct Answer - BCD `(A)Cl_(2_(g))to2Cl_((g)) " " ` randomness is increasing (B) In closed container `DeltaV=0` hence work cone is zero. There is no heat exchange . Hence `DeltaE=q+W=0` ( C)`DeltaG` will be zero only when equilibrium is reached. `:.` not a function of pressure (D)`DeltaG^@= - RT` In `K_(eq)` `:.` not a function of pressure

Discussion

20.	Space time in terms of molar federate and initial concentration is _________(a) τ = CA0 V0(b) τ = (frac{CA0}{FA0} )(c) τ = CA0 V(d) τ = (frac{VCA0}{FA0} )
Answer» Correct choice is (d) τ = (frac{VCA0}{FA0} ) Easy explanation: By definition, Space time = (frac{Volume , of , the , reactor}{Volumetric , feed , rate} = frac{V}{V0} ) and FA0 = CA0 V0, Therefore τ = (frac{VCA0}{FA0}. )

Discussion

21.	The design equation of PBR to determine the weight of the catalyst is ________(a) (frac{W}{FA0} = int_0^X )-rA dX(b) (frac{W}{FA0} = int_0^X frac{dX}{dV} )(c) W = FA0 (int_0^X frac{dX}{-rA} )(d) W = (int_0^X )FA0 dX
Answer» Right option is (c) W = FA0 (int_0^X frac{dX}{-rA} ) Explanation: The general mole balance for flow reactors is modified to get the above equation. The reaction rate is expressed in mol/s.g of catalyst in this equation.

Discussion

22.	The coefficients of absorption and reflection of the surface of a thin plate are \( 0.74 \) and \( 0.22 \) respectively. If \( 184 cal \) of radiant heat is incident on the surface of the plate, find the quantities of heat (i) absorbed (ii) reflected (iii) transmitted. (4 marks)
Answer» Given, the coefficient at absorption a = 0.74 The coefficient at reflection = r = 0.22 We know that a + r +t = 1 Where a = absorption coefficient r = reflection coefficient t = transmit coefficient 0.74 + 0.22 + t = 1 t = 1 - 0.96 t = 0.04 i) Quantities of heat absorption Q_a = a x Q Q_a = 0.74 x 184 Q_a = 136.16 cal ii) Quantities of heat reflection Q_r = r x Q Q_r = 0.22 x 184 Q_r = 40.08 cal iii) Quantities of heat transmitted Q_t = t x Q = 0.04 x 184 Q_t= 7.36 cal

22.

The coefficients of absorption and reflection of the surface of a thin plate are $ 0.74 $ and $ 0.22 $ respectively. If $ 184 cal $ of radiant heat is incident on the surface of the plate, find the quantities of heat (i) absorbed (ii) reflected (iii) transmitted. (4 marks)

Answer»

Given, the coefficient at absorption a = 0.74

The coefficient at reflection = r = 0.22

We know that

a + r +t = 1

Where

a = absorption coefficient

r = reflection coefficient

t = transmit coefficient

0.74 + 0.22 + t = 1

t = 1 - 0.96

t = 0.04

i) Quantities of heat absorption

Q_a = a x Q

Q_a = 0.74 x 184

Q_a = 136.16 cal

ii) Quantities of heat reflection

Q_r = r x Q

Q_r = 0.22 x 184

Q_r = 40.08 cal

iii) Quantities of heat transmitted

Q_t = t x Q

= 0.04 x 184

Q_t= 7.36 cal

Discussion

23.	What is cell potential or emf cell of a cell
Answer» The cell potential is the measure of potential difference between two half cells in an electrochemical cell. It is represented by the symbol $E_{c e l}$

Discussion

24.	10 kilograms of water are boiled at \( 373 K \), at a pressure of \( 1.013 \times 10^{5} N / m ^{2} \) and converted into steam. The specific latent heat of vaporization of water is \( 539 kcal / kg .1 L \) of water, on conversion into steam, occupies \( 1671 L \). The mechanical equivalent of heat is \( 4186 J / kcal \). Calculate (a) the energy supplied to the system (water)(b) the work done by the system(c) the change in the internal energy of the system.
Answer» Given P = 1.013 x 10⁵ N/ m² M = 10 kg L = 539 kcal/kg V₁ = 1 L V₂ = 1671 L (a) Q = ML = 10 x 10³ x 539 Q = 5390 kcal (b) Work done W = Pdv = 1.013 x 10⁵ (V₂- V₁) = 1.013 x 10⁵ (1671 -1) x 10^-3 W = \(\frac {1691.71 \times 10^2\,J}{4.2}\) W = 402.7 x 10² W = 40.27 kcal (c) dθ = du + dW du = dθ - dW = 5390 - 40.27 du = 5349.73 kcal

24.

10 kilograms of water are boiled at $ 373 K $, at a pressure of $ 1.013 \times 10^{5} N / m ^{2} $ and converted into steam. The specific latent heat of vaporization of water is $ 539 kcal / kg .1 L $ of water, on conversion into steam, occupies $ 1671 L $. The mechanical equivalent of heat is $ 4186 J / kcal $. Calculate (a) the energy supplied to the system (water)(b) the work done by the system(c) the change in the internal energy of the system.

Answer»

Given P = 1.013 x 10⁵ N/ m²

M = 10 kg

L = 539 kcal/kg

V₁ = 1 L

V₂ = 1671 L

(a) Q = ML

= 10 x 10³ x 539

Q = 5390 kcal

(b) Work done

W = Pdv

= 1.013 x 10⁵ (V₂- V₁)

= 1.013 x 10⁵ (1671 -1) x 10^-3

W = $\frac {1691.71 \times 10^2\,J}{4.2}$

W = 402.7 x 10²

W = 40.27 kcal

(c) dθ = du + dW

du = dθ - dW

= 5390 - 40.27

du = 5349.73 kcal

Discussion

25.	VI. Answer in briefly1. Define one calorie.2. Distinguish between linear, arial and superficial expansion.
Answer» 1. Define one calorie ? ans :- The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.

Discussion

26.	tell me one example about motor neurone
Answer» Motor neurons carry instructions from the brain along axons that stretch from the spinal cord to the muscles in the hands and feet. In diseases eg spinal muscular atrophy, motor neuron axons become damaged and degenerate, which means signals from the brain never reach the muscles and movement becomes impaired.

Discussion

27.	In young’s double slit experiment, fringes of certain.
Answer» Width are produced on the screen kept at a certain distance from the slits. When the screen is moved away from the slits by 0.1m, fringe width increases by 6 × 10^-5 m. The separation between the slits is 1mm. calculate the wavelength of the light used.

Discussion

28.	In which of the following equations the product formed has similar oxidation state for nitrogen?
Answer» (b) NO₂ and N₂O₄has + 4 oxidation state for nitrogen.

Discussion

29.	Write the experimental observations of photo electric effect.
Answer» The following one the other experimental results of photoelectric effect: 1. Photoelectric effect is an instantaneous process. 2. The minimum frequency is required to remove the electron from the metal surface is called threshold frequency. If the frequency of incident light is less than threshold frequency them their is no photoeletric effect. 3. The minimum energy required to remove the electron from the metal surface is called work function. ie. w = hυ₀ 4. The intensity of incident light is directly proportional to photoelectric current. 5. The kinetic energy of emitted electron is directly proportional to frequency of incident light. 6. A -ve potential is applied to anode plate just stop the photoelectric current is called stopping potential.

29.

Write the experimental observations of photo electric effect.

Answer»

The following one the other experimental results of photoelectric effect:

1. Photoelectric effect is an instantaneous process.

2. The minimum frequency is required to remove the electron from the metal surface is called threshold frequency. If the frequency of incident light is less than threshold frequency them their is no photoeletric effect.

3. The minimum energy required to remove the electron from the metal surface is called work function. ie. w = hυ₀

4. The intensity of incident light is directly proportional to photoelectric current.

5. The kinetic energy of emitted electron is directly proportional to frequency of incident light.

6. A -ve potential is applied to anode plate just stop the photoelectric current is called stopping potential.

Discussion

30.	Write any three experimental observations of photoelectirc effect.
Answer» 1. The photoelectric emission is an instantaneous process, even when incident radiation is exceedingly dim. 2. Above threshold frequency, the photo current is directly proportional to the intensity of incident radiation. 3. Above the threshold frequency, saturation current is proportional to the intensity of the incident radiation and stopping potential is independent of intensity. 4. There exists a certain minimum cut-off frequency called ‘threshold frequency’ below which no photo emission however intense the incident beam. 5. Above threshold frequency the kinetic energy of the photo electrons is directly proportional to the frequency of incident radiation and is independent of intensity.

30.

Write any three experimental observations of photoelectirc effect.

Answer»

1. The photoelectric emission is an instantaneous process, even when incident radiation is exceedingly dim.

2. Above threshold frequency, the photo current is directly proportional to the intensity of incident radiation.

3. Above the threshold frequency, saturation current is proportional to the intensity of the incident radiation and stopping potential is independent of intensity.

4. There exists a certain minimum cut-off frequency called ‘threshold frequency’ below which no photo emission however intense the incident beam.

5. Above threshold frequency the kinetic energy of the photo electrons is directly proportional to the frequency of incident radiation and is independent of intensity.

Discussion

31.	Light from any ordinary source (such as a flame) is usually: A. unpolarized B. plane polarized C. circularly polarized D. elliptically polarized E. monochromatic
Answer» A. unpolarized

Discussion

32.	The electric field in unpolarized light: A. has no direction at any time B. rotates rapidly C. is always parallel to the direction of propagation D. changes direction randomly and often E. remains along the same line but reverses direction randomly and often
Answer» D. changes direction randomly and often

Discussion

33.	As used in the laws of reflection and refraction, the “normal” direction is: A. any convenient direction B. tangent to the interface C. along the incident ray D. perpendicular to the electric field vector of the lightE. perpendicular to the interface
Answer» E. perpendicular to the interface

Discussion

34.	The index of refraction of a substance is: A. the speed of light in the substance B. the angle of refraction C. the angle of incidence D. the speed of light in vacuum divided by the speed of light in the substance E. measured in radians
Answer» D. the speed of light in vacuum divided by the speed of light in the substance

Discussion

35.	When an electromagnetic wave meets a reflecting surface, the direction taken by the reflected wave is determined by: A. the material of the reflecting surface B. the angle of incidence C. the index of the medium D. the intensity of the wave E. the wavelength
Answer» B. the angle of incidence

Discussion

36.	A particle of mass `m` is rotating in a plane in circular path of radius `r`. Its angular momentum is `L`. The centripetal force acting on the particle isA. `L^(2)//mr`B. `L^(2)m//r^(2)`C. `L^(2)//m^(2)r^(2)`D. `L^(2)//mr^(3)`
Answer» Correct Answer - D A particle ………….. `F = (Mv^(2))/(r ) = L^(2)//mr^(3)`

Discussion

37.	1. Define simple harmonic motion for a particle moving in a straight line.2. Use your definition to explain how simple harmonic motion can be represented by the equation.3. Show that the above equation is dimensionally correct.4. A mechanical system is known to perform simple harmonic motion. What quantity must be measured in order to determine frequency for the system?
Answer» 1. A periodic motion in which acceleration is directly proportional to displacement and opposite in direction is called simple harmonic oscillation. 2. Mathematically, a simple harmonic oscillation can be expressed as y = a sin wt (or) y = a cos wt 3. y = a sin w t Sin wt has no dimension. Hence we need to consider dimension of ‘a’ only, ie, y = a, L = L 4. Its period is determined.

37.

1. Define simple harmonic motion for a particle moving in a straight line.2. Use your definition to explain how simple harmonic motion can be represented by the equation.3. Show that the above equation is dimensionally correct.4. A mechanical system is known to perform simple harmonic motion. What quantity must be measured in order to determine frequency for the system?

Answer»

1. A periodic motion in which acceleration is directly proportional to displacement and opposite in direction is called simple harmonic oscillation.

2. Mathematically, a simple harmonic oscillation can be expressed as

y = a sin wt (or) y = a cos wt

3. y = a sin w t

Sin wt has no dimension. Hence we need to consider dimension of ‘a’ only, ie, y = a, L = L

4. Its period is determined.

Discussion

38.	Both the strings, shown in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of transverse wave in the string `AB` is `v_(1)` and in `CD` it is `v_(2)`, the `v_(1)//v_(2)` is A. 1B. 2C. `sqrt2`D. `(1)/(sqrt2)`
Answer» Correct Answer - D `v=sqrt((T)/(u))` [T = tension forse] [`mu` = mass per unit length]

Discussion

39.	The LCM of 3a, 15ab is ______.1. 15ab2. 14ab3. Ab4. None of these
Answer» Correct Answer - Option 1 : 15ab Given: The LCM of 3a, 15ab Calculation: The prime factor of 3a, 15ab = 3 , 5, a, b The LCM = 3 × 5 × a × b = 15ab ∴ The LCM of 3a, 15ab is 15ab. The Least Common Multiple ( LCM ) is also referred to as the Lowest Common Multiple ( LCM ) and Least Common Divisor ( LCD) . For two integers a and b, denoted LCM(a, b), the LCM is the smallest positive integer that is evenly divisible by both a and b.

39.

The LCM of 3a, 15ab is ______.1. 15ab2. 14ab3. Ab4. None of these

Answer» Correct Answer - Option 1 : 15ab

Given:

The LCM of 3a, 15ab

Calculation:

The prime factor of 3a, 15ab = 3 , 5, a, b

The LCM = 3 × 5 × a × b = 15ab

∴ The LCM of 3a, 15ab is 15ab.

The Least Common Multiple ( LCM ) is also referred to as the Lowest Common Multiple ( LCM ) and Least Common Divisor ( LCD) . For two integers a and b, denoted LCM(a, b), the LCM is the smallest positive integer that is evenly divisible by both a and b.

Discussion

40.	What is the maximum and minimum possible resistance which can be determined using the `PO` Box shown in above .A. `1111 kOmega, 0.1Omega`B. `1111 kOmega, 0.01Omega`C. `1111 kOmega, 0.001Omega`D. None of these
Answer» `X=(Q)/(P)RrArr(X)_(max)=((Q)_(max))/((P)_(max))(R)_(max)=(1000)/(10)(11110)=1111kOmega` `(X)_(min)=((Q)_(min))/((P)_(max))( R)_(min)=(10)/(1000)(10)/(10000)(1)=0.01Omega` .

Discussion

41.	In a region there exists a magnetic field `B_(0)` along positive x-axis. A metallic wire of length 2a, one side along x-axis and one side parallel of y-axis is rotates about y-axis with a angular velocity. Then at the instant shown A. Potential difference across PQ is 0B. Potential difference across PQ is `1/2 B_(0)omega a^(2)`C. Potential difference across QR is `1/2 B_(0)omega a^(2)`.D. Potential difference across QR is `B_(0)omega a^(2)`.
Answer» Correct Answer - A::D PQ is parallel to magnitude field, hence no emf is induced in it. QR is `_\|_^(r)` perpendicular to magnetic field. Induced emf =`Bv1=B_(0)omega a^(2)`.

Discussion

42.	Dimensional formula of stress is same as that of the pressure. Is it true or false?
Answer» Dimensional formula of stress is the same as that of the pressure is true.

Discussion

43.	Define strain..
Answer» Strain produced in a body is defined as ratio of change in dimension to the original dimension.

Discussion

44.	Stress is a vector quantity. Is it true or false?
Answer» Stress is a vector quantity is false.

Discussion

45.	When work done on a body equals the change in its kinetic energy, this principle is known as
Answer» Work done on body equals to change in its kinetic energy is known as work-energy principle. Work- energy principle

Discussion

46.	Colored TV has approximate power of
Answer» Answer: 120 watt

Discussion

47.	A man moves on his motorbike with `54 km//h` and then takes a U-turn and containues to move with same velocity The time of U-turn is `10 s`. Find the magnitude of average acceleration during U-turn.
Answer» Correct Answer - B `54km//h=54xx(5)/(18)=15m//s` `lt a gt=(15-(-15))/(10)=3m//s^(2)`

Discussion

48.	A vernier callipers has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are ofA. 0.5mmB. 1mmC. 2mmD. 1/4mm
Answer» `20VSD=19MSD` `1VSD=(19)/(20)MSD` `L.C.=1MSD-1VSD=1MSD-(19)/(20)MSD` `L.C=(1)/(20)MSD` `rArr0.1=(1)/(290)MSDrArrMSD=2mm`.

Discussion

49.	A vernier callipers has 20 divisions on the vernier scale which coincide with 19 divisions on the main scale. The least count of the instrument is 0.1 mm. The main scale divisions are ofA. 0.2mmB. 0.5mmC. 1.0mmD. 2.0mm
Answer» N VSD=(N-1)MSD `20 VSD=19MSD therefore MSK=(20)/(19)VSD` `L.C.=1 MSD+1VSD=1MSK=(19)/(20)MSK` `L.C.=(1-(19)/(20))MSD` `L.C.=(1)/(20)xxMSD` `0.1mm=(1)/(20)xxMSD` `therefore MSD=0.1mmxx20` `MSD=2mm`

Discussion

50.

Two point charges +9e and +1e are kept at a distance of 16 cm from each other. At what point between these charges should a third charge q be placed so that it remains in equilibrium.

Answer»

Solution:

Given charges: +e and +9e

Let the third chare q be placed at a distance x from e and hence its distance from 9e is 16-x cm

For the system to remain in equilibrium,

F₁ = F₂

Therefore,

F₁ = K(e q)/x²

F₂ = K(9e)/(16-x)²

Equating F₁ and F₂ , we get,

9/x² = 1/(16-x)²

Taking square root on both sides:

3/x = 1/(16-x)

Solving for x, we get,

x = 12cm

Hence, the charge q should be placed at 12 cm from +9e.

Discussion

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