100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Two cells of emf 3 V and 4 V and internal resistance 1 omega and 2 omega respectively are connected in parallel so as to send the current in the same direction through an external resistance of 5Q. Find the potential difference across 5Q resistor.
Answer» Given : r₁ = 1Ω r₂= 2 Ω E₁ = 3v E₂ = 4v Total equivalent internal resistance \(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{r_1}\) + \(\cfrac{1}{r_2}\) \(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{1}\) + \(\cfrac{1}{2}\) \(\cfrac{1}{r_{eq}}\) = \(\cfrac{3}{2}\) r_eq = \(\cfrac{2}{3}\)Ω Equivalent emf of cell in parallel connection E_{eq = \(\left[\cfrac{E_1}{r_1}+\cfrac{E_2}{r_2}\right]r_{eq}\)} E_{eq = _{\(\left[\cfrac{3}{1}+\cfrac{4}{2}\right]\times \cfrac23\)}} E_{eq =}_{_{\(\left[\cfrac{6+4}2\right]\times \cfrac23\)}} E_{eq = 5 x}\(\cfrac{2}{3}\) E_{eq = 3.3 v} potential difference across 5 Ω resistor E = _{_{\(\left[\cfrac{R}{R+r_{eq}}\right]\)}}E_eq E = \(\left[\cfrac{5}{5+\cfrac23}\right]\) x 3.3 = \(\cfrac{15}{17}\) x 3.3 = 2.9 v

1.

Two cells of emf 3 V and 4 V and internal resistance 1 omega and 2 omega respectively are connected in parallel so as to send the current in the same direction through an external resistance of 5Q. Find the potential difference across 5Q resistor.

Answer»

Given : r₁ = 1Ω

r₂= 2 Ω

E₁ = 3v

E₂ = 4v

Total equivalent internal resistance

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{r_1}\) + \(\cfrac{1}{r_2}\)

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{1}{1}\) + \(\cfrac{1}{2}\)

\(\cfrac{1}{r_{eq}}\) = \(\cfrac{3}{2}\)

r_eq = \(\cfrac{2}{3}\)Ω

Equivalent emf of cell in parallel connection

E_{eq = \(\left[\cfrac{E_1}{r_1}+\cfrac{E_2}{r_2}\right]r_{eq}\)}

E_{eq = _{\(\left[\cfrac{3}{1}+\cfrac{4}{2}\right]\times \cfrac23\)}}

E_{eq =}_{_{\(\left[\cfrac{6+4}2\right]\times \cfrac23\)}}

E_{eq = 5 x}\(\cfrac{2}{3}\)

E_{eq = 3.3 v}

potential difference across 5 Ω resistor

E = _{_{\(\left[\cfrac{R}{R+r_{eq}}\right]\)}}E_eq

E = \(\left[\cfrac{5}{5+\cfrac23}\right]\) x 3.3

= \(\cfrac{15}{17}\) x 3.3

= 2.9 v

Discussion

2.	A cyclist starts from the centre O of a circular park of radius 1km, reaches the edge P of the park, then cycles along the PQ cicumference and returns to the centre along OQ as shown in fig. If the round trip taken ten minute, the net displacement and average speed of the cylists (in kilometer and kinetic per hour) is A. 0, 1B. `(pi +4)/(2), 0`C. `21.4, (pi + 4)/(2)`D. 0, 21.4
Answer» Correct Answer - 4 Net disp.`=0` and `V_(avg)=(2+(pir)/(2))/(((1)/(6)))` `=21.4 (km)/(hr)`

Discussion

3.	A body travels uniformly a distance of `(10.0 +- 0.5 m)` in a time `(2.0 +- 0.1)` sec. The velocity of the body within error limits is :A. `(5.0 +- 0.6)` m/sB. `(5.0 +- 0.5)` m/sC. `(5.0 +- 0.05)` m/sD. `(5.0 +- 1.0)` m/s
Answer» Correct Answer - 2 `v=("distance")/("time")=((10.0+-0.5)m)/((2.0+-0.1)s)` `"distance"=(10.0+-(0.5)/(10.0)xx100%)` `=(10.0+-5%)` `"time"=(2.0+-(0.1)/(2.0)xx100%)` `=(2.0+-5%)` `"velocity"=5.0+-10%=5.0+-(5.0xx10%)`

Discussion

4.	The four capacitors, each of `25 muF` are connected as shown in figure. The DC voltmeter reads 200V. The charge on each plate of capacitor is A. `pm2xx10^(-3)C`B. `pm5xx10^(-3)C`C. `pm2xx10^(-2)C`D. `pm5xx10^(-2)C`
Answer» Correct Answer - B `Q=CV` `Q=25xx10^(-6)xx200` `Q=pm5xx10^(-3)C`

Discussion

5.	A body of mass 5 kg has momentum of 10 kg m/s. When a force of 0.2 N is applied on it for 10 seconds. What is the change in its kinetic energy (1) 1.1 J(2) 2.2 J (3) 3.3 J (4) 4.4 J
Answer» Answer is (4) 4.4 J u = p/m = 10/5 = 2m/s, a = (0.2/5) m/s². ν = u + at = 2 + (0.2/5) x 10 = 2.4 m/s². ∆K = (1/2) mν² - 1/2 mu² = (1/2) x 5 x (2.4)² – (1/2) x 5 x 2² = 4.4 J

5.

A body of mass 5 kg has momentum of 10 kg m/s. When a force of 0.2 N is applied on it for 10 seconds. What is the change in its kinetic energy (1) 1.1 J(2) 2.2 J (3) 3.3 J (4) 4.4 J

Answer»

Answer is (4) 4.4 J

u = p/m = 10/5 = 2m/s, a = (0.2/5) m/s².

ν = u + at = 2 + (0.2/5) x 10 = 2.4 m/s².

∆K = (1/2) mν² - 1/2 mu² = (1/2) x 5 x (2.4)² – (1/2) x 5 x 2² = 4.4 J

Discussion

6.	A particle of mass ‘m’ moving with velocity ‘v’ collides in elastically with a stationary particle of mass ‘2m’. The speed of the system after collision will be – (1) ν/2(2) 3ν (3) ν/3(4) 3ν
Answer» Answer is (3) ν/3 m₁ν₁ + 0 = (m₁+ m₂)ν’ ⇒ ν’ = mν/(m + 2m) = ν/3

6.

A particle of mass ‘m’ moving with velocity ‘v’ collides in elastically with a stationary particle of mass ‘2m’. The speed of the system after collision will be – (1) ν/2(2) 3ν (3) ν/3(4) 3ν

Answer»

Answer is (3) ν/3

m₁ν₁ + 0 = (m₁+ m₂)ν’

⇒ ν’ = mν/(m + 2m) = ν/3

Discussion

7.	A particle has displacement of 12 m towards east and 5m towards north then 6 m vertically upward. The sum of these displacements is (1) 12. (2) 10.04 m. (3) 14.31 m (4) none of these
Answer» Answer is (3) 14.31 m S = S1 + S2 + S3 S = √(S₁² + S₂² + S₃²) ⇒ S = √(12² + 5²+ 6²) = 14.31 m

7.

A particle has displacement of 12 m towards east and 5m towards north then 6 m vertically upward. The sum of these displacements is (1) 12. (2) 10.04 m. (3) 14.31 m (4) none of these

Answer»

Answer is (3) 14.31 m

S = S1 + S2 + S3

S = √(S₁² + S₂² + S₃²)

⇒ S = √(12² + 5²+ 6²) = 14.31 m

Discussion

8.	A doubly ionized helium ion and a hydrogen ion are accelerated through the same potential. The ratio of the speeds of helium and hydrogen ions is (1) 1 : 2 (2) 2 :1 (3) 1 : √2 (4) √2 : 1
Answer» Answer is (3) 1 : √2 E = (1/2)mν² = qV ⇒ ν = √(2qV/m) ⇒ ν α √(q/m) ν_He/ν_H = √(2/1) x (1/4) = 1/√2

8.

A doubly ionized helium ion and a hydrogen ion are accelerated through the same potential. The ratio of the speeds of helium and hydrogen ions is (1) 1 : 2 (2) 2 :1 (3) 1 : √2 (4) √2 : 1

Answer»

Answer is (3) 1 : √2

E = (1/2)mν² = qV

⇒ ν = √(2qV/m)

⇒ ν α √(q/m)

ν_He/ν_H = √(2/1) x (1/4) = 1/√2

Discussion

9.	A pendulum bob has a speed of `3ms^(-1)` at its lowest position. The pendulum is `0.5` m long. The speed of the bob, when string makes an angle of `60^(@)` to the vertical is `("take, g"=10ms^(-1))`A. `(3)/(2)m//s`B. 2m/sC. `(1)/(2)m//s`D. 3m/s
Answer» Correct Answer - B According to the law of conservation of energy `TE_(at""L) = TE_(at""h)` `(1)/(2)mv^(2) = (1)/(2)"mu"^(2) + mgh` `v^(2) = u^(2) + 2gh or u^(2) = v^(2) - 2gh` ` u^(2) = 9 - 2 xx 10 xx (1)/(2)(1-(1)/(2))` `u^(2) = 9 - 5 = 4` ` u=2 ms^(-1)`

Discussion

10.	`20%` radioactive sample decay in time t. How much sample decay in time 2t ?A. `40%`B. `36%`C. `44%`D. `80%`
Answer» Correct Answer - B `20%` decay in timet means `80%` remaining , `(N)/(N_(0))= (80)/(100) =(8)/(10)` In time `2t ((N)/(N_(0))) = ((80)/(100))^(2) = .64` `rArr 64%` remaining means `36%` deecay.

Discussion

11.	The intensity of light pulse travelling in an optical fiber decreases according to the relation `I=I_(0)e^(-alpha x)` . The intensity of light is reduced to `20%` of its initial value after a distance x equal toA. In`((1)/(alpha))`B. In`(alpha)`C. `(("In"5))/(alpha)`D. In `((5)/(alpha))`
Answer» Correct Answer - C `(I)/(I_(0)) = e^(-alpha x) = (20)/(100) = e^(alpha x) = 5` `ax = log_(e) 5, x=(("In" 5))/(alpha)` OR `I=I_(0)e^(-alpha x)` `0.2 I_(0) = I_(0) e^(-alpha x) ` `(2)/(10) = e^(-alpha x) rArr (10)/(2) = e^(alpha x)` `l n(5) = dx` `x=(l n(5)J)/(alpha) `

Discussion

12.	The momentum of a photon is `3.3xx10^(-29)`kg-m/s. Its frequency will beA. `3xx10^(3)Hz`B. `6xx103Hz`C. `7.5xx10^(12)Hz`D. `1.5 xx 10^(13) Hz`
Answer» Correct Answer - D `p=(hv)/(c ) ` `rArr = (pc)/(h) = (3.3 xx 10^(-19) xx 3 xx 10^(8))/(6.6xx10^(-34))` `=1.5 xx 10^(13)Hz`

Discussion

13.	The perio of oscillation of a magnet in vibration magnetometer is 2 sec. The period of oscillation of a magnet whosr magnetic moment is four times that of the first magnet isA. 8sB. 4sC. 1sD. 0.5 s
Answer» Correct Answer - C The tme period `T=2pi sqrt(I//MB_(H))` Thus `(T_(2))/(T_(1)) = sqrt((M_(1))/(M_(2)))=sqrt((1)/(4)) =(1)/(2)` `or T_(2) = 2 xx I//2` = 1 sec the answer is (3) 1 sec

Discussion

14.	Which of the following is paramagnetic and HOMO is gerade molecular orbitalA. `O_(2)`B. `O_(2)^(2-)`C. `B_(2)`D. `C_(2)`
Answer» Correct Answer - A

Discussion

15.	Which of the following species have two types of C-C covalent bond length dataA. diamondB. graphiteC. FullereneD. None of these
Answer» Correct Answer - C

Discussion

16.	The correct order of size is :A. `Sc lt V lt Cr`B. `Sc lt Y = La`C. `Ni lt Cu lt Zn`D. `Cu lt Au lt Ag`
Answer» Correct Answer - C `(A) Scgt V gt Cr " "(B) Sc lt Ylt La" " (C ) Cu ltAg le Au`

Discussion

17.	Which of the following compound will have only primary and secondary carbonA. PropaneB. 2, 2, 3-trimethylpentaneC. 2 -methylpropaneD. 1 -bromopropane
Answer» Correct Answer - A::D `(A) Coverset(1^(@))(H_(3))-overset(2^(@))(CH_(2))-overset(1^(@))(CH_(3))" "(D)Coverset(1^(@))(H_(3))-overset(2^(@))CH_(2)-underset(Br)underset(\|)overset(1^(@))(CH_(2))`

Discussion

18.	Two aqueous solutions of `AgNO_(3)` and NaCl are mixed. Which of the following diagrams best represents the mixture after reaction?A. B. C. D.
Answer» Correct Answer - C `AqNO_(3)(aq)+NaCl(aq)rarrAgCl(s)+NaNO_(3)(aq)`

Discussion

19.	For 0.2 mol `C_(3)H_(4)` and 0.5 mol `C_(2)H_(6)` gas mixture? Mass of mixture is 23 g It includes 1.6 mol C atom It includes `0.7 xx 6.022 xx 10^(23)` total molecules Which one of the following option is correct?A. I , II onlyB. II, III onlyC. I, II, IIID. I, III only
Answer» Correct Answer - C Mass of mixture `= 0.2 xx40+0.5 xx30=23 gm` Mole of c-atom `=0.2xx3+0.5xx2=1.6` mole Number of molecule `=(0.2 +0.5)N_(A)`

Discussion

20.	The number of gerade atomic orbitals present in `Cu^(+)` (z = 29)A. 10B. 14C. 8D. 6
Answer» Correct Answer - C `Cu^(+)underset(1)ubrace(1s^(2))underset(1)ubrace(2s^(2))2p^(6)underset(1)ubrace(3s^(2))3p^(6)underset(5)ubrace(3d^(10))`

Discussion

21.	Which of the following set of Quantum numbers is not possible?A. n=4,l=3,m2,s=+1/2B. n=4,l=2,m=+2,s=-1/2C. n=4,l=2,m=-2,s=+1/2D. n=4,l=1,m=-2,s=-1/2
Answer» If l-1 then permissible values of m=-1,0,+1

Discussion

22.	Final the total number of species having two unparired electron from the following species, `Fe^(2+),Cr,Cr^(3+),Ti^(2+),V^(3+)`
Answer» `{:(Fe^(+2),[Ar],3d^(5)),(Cr,[Ar],3d^(5)4s^(1)),(Cr^(+3),[Ar],ed^(3)),(Ti^(+2),[Ar],3d^(2)),(Mn^(+2),[Ar],3d^(5)),(V^(+3),[Ar],3d^(2)):}`

Discussion

23.	Calculate the total number of p-orbitals electrons present in `Cu(29)` atoms. [Divide answer by 2]
Answer» `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)4s^(1)3d^(10)`

Discussion

24.	Which of the following has maximum unpaired electrons ?A. `Fe^(3+)`B. `Fe^(2+)`C. `Mn^(3+)`D. `So^(3+)`
Answer» `Fe^(3+)=[Ar]4s^(@)3d^(5)`

Discussion

25.	The correct option regarding size of orbitals is:A. `3pgt4pgt5p`B. `3plt4p=5p`C. `3plt4plt5p`D. `3p=4p=5p`
Answer» Value of n `uparrow` size of orbital `uparrow`

Discussion

26.	Find minimum value of atomic number for which all orbitals of n=3 are filled. [Divide your answer by 10]
Answer» `1s^(2)2s^(2)2p^(6)4s^(2)3d^(10)` Atomic no=30 `=(30)/(10)=3`

Discussion

27.	Which of the following energy level can not exist according to quantum theory?A. 3fB. 5fC. 5hD. 6h
Answer» `lrarrs p d f g h` 0 1 2 3 4 5 n= always greater than l `3frarrn=3,l=3,` this is not possible `5hrarrn=4,l=5` This is not possible

Discussion

28.	A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration towards the righ. Pressure is i. maximum and ii. minimum at A. i. B ii DB. i C ii DC. i B ii CD. i B ii A
Answer» Correct Answer - A `P_A=P_D+rhoal,P_B=P_A+rhogh, P_C=P_D+rhogh` `P_B=P_D+rhogh+rhoal`

Discussion

29.	A dimensionless quantity(a) never has a unit, (b) always has a unit,(c) may have a unit, (d) does not exist.
Answer» (c) may have a unit, Explanation: Dimensionless quantities may have units.

29.

A dimensionless quantity(a) never has a unit, (b) always has a unit,(c) may have a unit, (d) does not exist.

Answer»

(c) may have a unit,

Explanation:

Dimensionless quantities may have units.

Discussion

30.	A metallic wire is folded to form a square loop of side `a`. It carries a current `i` and is kept perpendicular to the region of uniform magnetic field `B`. If the shape of the loop is changed from square to an equilateral triangle without changing the length of the wire and current. The amount of work done in doing so isA. `Bia^(2)(1-(4sqrt(3))/(9))`B. `Bia^(2)(1-(sqrt(3))/(9))`C. `(2)/(3)Bia^(2)`D. Zero
Answer» emf. Induced in coil `epsilon=-(Deltaphi)/(Deltat)=-((phi_(2)-phi_(1)))/(t_(2)-t_(1))` `phi_(2)=B.A_(2)-B.(4a)/(2xx3)xx(4a)/(3)xxsin60^(@)=(4sqrt(3))/(9)Ba^(2)` `phi_(1)=Ba^(2)` Work done `W=epsilonI Deltat=Ba^(2)(1-(4sqrt(3))/(9))i`

Discussion

31.	A current flows through a rectangular conductor in the presence of uniform magnetic field B pointing out of the page as shown. Then the potential difference `V_(P)-V_(Q)` is equal to (assume charge carriers in the conductor to be positively charged moving with a drift velocity of v ) A. `Bvb`B. `-Bvb`C. `Bve`D. `-Bve`
Answer» Correct Answer - C `\|epsilon\|=B lv` where l is the edge perpendicular to both B and `vecv` i.e.c. `therefore \|epsilon\|=B v c`. Now by right hand thumb rule magnetic force an a positive charge moving towards right is in down wrd direction Hence end P will be positive. `therefore V_(P)-V_(!)` is positive `rArr epsilon=+Bvl`.

Discussion

32.	A charged particle of mass `m` & charge `q` enters a zone of uniform magnetic field `B` with a velocity `v` making an angle `theta` with the boundary of the zone, having width `d`, when the particle penetrates half-way into the zone, the change in energy of the particel is A. `(mv^(2))/(2B)sintheta`B. `(Bqd)/(mv^(2))sintheta`C. zeroD. `(mvd)/(qB)sintheta`
Answer» As magnetic force is always perpendicular to velocity. No work is done by force.

Discussion

33.	The magnetic field strength at a point `P` distant `r` due to an infinite straight wire carrying a current `i` is as shown in the figure A. `mu_(0)`B. `mu_(0)i//2sqrt(2)r`C. `(mu_(0)i//sqrt(2)pir)`D. `(mu_(0)i)/(4pir)[2+sqrt(2)]`
Answer» `B_("net")=2(mu_(0)I)/(4pir)[sin90^(@)+sin45^(@)]o.`

Discussion

34.	Only circular part of the wire shown in the figure has resistance `sigam` per unit length. The radii of the circles are `r` and `R` and `/_POQ=alpha` (in radian). If a potential difference `V_(P)-V_(Q)=V` is applied across the point `P` and `Q`, the magnetic field at point `O` is A. `(mu_(0))/(4pi)(V)/(sigma)[(2pi-alpha)/(r^(2))-(alpha)/(R^(2))]`B. `(mu_(0)V)/(4pi)[(alpha)/(r^(2))-(2pi-alpha)/(R^(2))]`C. `(mu_(0))/(4pi)(V)/(sigma)[(1)/(r^(2))-(1)/(R^(2))]`D. `(mu_(0))/(4pi)(sigma)/(V)[(1)/(r^(2))-(1)/(R^(2))]`
Answer» `i_(R )=(V)/(alphaRsigma)("cc"w)i_(r)=(V)/((2pi-alpha)rsigma)(cw)`

Discussion

35.	The magnetic field at the centre of the circular loop as shown in Fig. when a single wire is bent to form a circular loop and also extends to form straight sections is A. `(mu_(0)I)/(2R)otimes`B. `(mu_(0)I)/(2R)(1+(1)/(pisqrt(2)))odot`C. `(mu_(0)I)/(2R)(1-(1)/(pisqrt(2)))odot`D. `(mu_(0)I)/(2R)(1+(1)/(pisqrt(2)))otimes`
Answer» `B_(AB)=(mu_(0)I)/(4piR)[sin90^(@)+sin(-45^(@))]ox` `B_(BC)=(mu_(0)I)/(4piR)[sin90^(@)+sin45^(@)]o.` `B_(0)=B_(AB)+B_(BC)+B_("circular")`

Discussion

36.	A steady current is flowing in a circular coil of radius R, made up of a thin conducting wire. The magnetic field at the centre of the loop is `B_L`. Now, a circular loop of radius `R//n` is made from the same wire without changing its length, by unfounding and refolding the loop, and the same current is passed through it. If new magnetic field at the centre of the coil is `B_C`, then the ratio `B_L//B_C` isA. `1 : n^(2)`B. `1 : L`C. `n : n^(2)`D. `n : n^(1//2)`
Answer» `vecB_(0)=(mu_(0)ni)/(2R)` (for circular ring with `n` loops) `:. ` Length remains constant `:. r_(2)=R//n`

Discussion

37.	Concept of heat
Answer» Heat may be defined as energy in transit. Word heat is used only if there is a transfer of energy from one thermodynamic system to the another. When two systems at different temperatures are kept in contact with each other than after some time temperatures of both the systems become equal and this phenomenon can be described by saying that energy has flown from one system to another. This flow of energy from one system to another on account of temperature difference is called heat transfer. Flow of heat is a non-mechanical mode of energy transfer. Heat flow depends not only on initial and find states but also on path it's.

Discussion

38.	We have seen that a gamma-ray does of `3 Gy` is lethal to half the people exposed to it. If the equivalnet energy were absorbed as heat, what rise in body temperature would result ?A. `300 muK`B. `700 muK`C. `455 muK`D. `390 muK`
Answer» Correct Answer - B We can relate an absorbed energy `Q` and the resulting temperature increase `DeltaT` with relation `Q = cmDeltaT`. In that equation, `m` is the mass of the materical obsorbing the energy and `c` is the specific heat of that material. An absorbed dose of `3Gy` correconds to an absorbed energy per unit mass of `3J//kg`. Let us assume that `c` the specific heat of human body, is the same as that of water, `4180 J//kg K`. then we find that `DeltaT = (Q//m)/(c) = (3)/(4180) = 7.2 xx 10^(-4) K~~ 700 muK` Obviously the damage done by ionizing radiation has nothing to do with thermal heating. The harmful effects arise because the radiation damages `DNA` and thus interferes with the normal functioning of tissues in which it is obsorbed.

Discussion

39.	Which of the following phenomena taken place when a monochromatic light is incident on a prism?(a) Dispersion(b) Deviation(c) Interference(d)All of the above
Answer» Correct answer is (b) Deviation

Discussion

40.	Write Brewster law of polarisation of light.
Answer» Brewster's law of polarisation of light: It states that when light is incident at polarising angle (θ_p) at the interface of a transparent medium, the refractive index of the medium w.r.t. the surrounding medium is equal to the tangent of the tangent of the polarising angle ¹μ₂ = tan θ_p.

Discussion

41.	Conductivity of a semiconuctor increases when a radiation of wavelength is less than `2480 nm` is incident on it. The forbidden gap isA. `0.5 J`B. `0.5 eV`C. `1 eV`D. `2 eV`
Answer» Correct Answer - B `"energy"=[(12400)/(lambda "in" A^(@))]eV`

Discussion

42.	An aeroplane needs to reach a speed of 324 km/hr before it can take off. It starts from rest and the engine of aeroplane produces a constant acceleration. If it spends 30 s on the runway then the minimum required runway length is -A. 1350 mB. 675 mC. 337.5 mD. None of these
Answer» Correct Answer - A `S=(U+V)/(2)xxt, S=(0+90)/(2)xx30,S=1350m`

Discussion

43.	A ball moving with a velocity `v` hits a massive wall moving towards the ball with a velocity a. An elastic impact lasts for time `/_ `A. b, cB. a, dC. b, dD. c, d
Answer» Correct Answer - A Velocity of ball after collision is `V_("bal")~=-(v+2u)and v_("wall")~=u` Since, `F_("ball")=((Deltap)/(Deltat))_("ball")=(-m(v+2u)-mv)/(Deltat)` `rArr" "F_("ball")=(-2m(v+u))/(Deltat)rArr\|F_("ball")\|=(2m(v+u))/(Deltat)` Increase in K.E. of ball is given by `DeltaK=(1)/(2)m(v+2u)^(2)-(1)/(2)mv^(2), DeltaK=2 m u(u+v)`

Discussion

44.	When a long spring is stretched by 2cm, its potential energy is U. If the spring is stretched by 10cm, the potential energy stored in it will beA. 25UB. U/5C. 5UD. 10U
Answer» Correct Answer - A `U prop x^(2)` `(U_(2))/(U_(1))=(x_(2)^(2))/(x_(1)^(2))rArr (U_(2))/(U)=10(2)^(2)=25 rArr U_(2)=25U_(1)=25U`

Discussion

45.	Three-pin plug is used in the circuit of power devices like electric iron.a. In three pin plug, what each pin indicate?b. Which part of the device is connected to the earth pin?
Answer» a. E-Earth line, N-Neutral line, P-Phase line. b. To the body of the device.

45.

Three-pin plug is used in the circuit of power devices like electric iron.a. In three pin plug, what each pin indicate?b. Which part of the device is connected to the earth pin?

Answer»

a. E-Earth line, N-Neutral line, P-Phase line.

b. To the body of the device.

Discussion

46.	Why do we say that metallic devices should be earthed?
Answer» Chances of electric shock are” more when metallic devices are used. When metallic devices are earthed, chances of electric shock can be prevented.

Discussion

47.	What is the role of earthing wire in a household circuit?
Answer» The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If at all the body comes into contact with an electric connection, electricity flows to the; earth through the earth wire. The flow of current to the earth through a circuit of low resistance increases the current. As a result heat generated in the fuse wire increases and the circuit gets broken. This ensures the safety of instrument and the person handling it.

47.

What is the role of earthing wire in a household circuit?

Answer»

The pin E of a three-pin plug comes into contact with the earth line. This pin is now connected to the body of the appliance. If at all the body comes into contact with an electric connection, electricity flows to the; earth through the earth wire.

The flow of current to the earth through a circuit of low resistance increases the current. As a result heat generated in the fuse wire increases and the circuit gets broken. This ensures the safety of instrument and the person handling it.

Discussion

48.	Which are the devices connected in series in a household circuit?
Answer» Fan, Switch, Regulator, etc.

Discussion

49.	A flowchart of art household wiring is given in the diagram below. Fix the suitable devices in the space provided by picking from the given list.(ELCB, MCB, Main switch, Watt-hour meter, Three-pin plug)
Answer» (A) Main switch. (B) ELCB. (C) MCB. (D) Three-pin plug.

49.

A flowchart of art household wiring is given in the diagram below. Fix the suitable devices in the space provided by picking from the given list.(ELCB, MCB, Main switch, Watt-hour meter, Three-pin plug)

Answer»

(A) Main switch.

(B) ELCB.

(C) MCB.

(D) Three-pin plug.

Discussion

50.	Construct such a circuit in aplywood as shown in Fig. The list of materials required and the number of items are given in the table.
Answer» Main switch. Fuse. MCB. Main fuse. Switchboard. ELCB. Fan. Bulb.

Discussion

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Construct such a circuit in aplywood as shown in Fig. The list of materials required and the number of items are given in the table.