InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In insulators (CB is conduction band and VB is valence band )A. VB is partially filled with electronsB. CB is partially filled with electronsC. CB is empty and VB is filled with electronsD. CB is filled with electrons and VB is empty |
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Answer» Correct Answer - C In insulators conduction band is empty and valence band is filled with electrons . |
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| 2. |
A body is moved along a straight line by a machine delivering constant power . The distance moved by the body is time `t` is proptional toA. `t^(1//2)`B. `t^(2//3)`C. `t`D. `t^(3//2)` |
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Answer» Correct Answer - D Power , P = Force `xx` Velocity = `F xx v` `=ma xx v` = constant So , va = constant `v""(dv)/(dt)` = constant Integrating it , we have `(v^(2))/(2) = Ct` or ` v = sqrt(2CT) = C_(1) t^(1//2)` or `(dx)/(dt) = C_(1)t^(1//2)` Integrating it , we get `x = C_2 t^(3//2)` So , `x prop t^3//2 ` |
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| 3. |
In cyclotron for a given magnet radius of the semicircle traced by positive ion is directly proportional to (where v= velocity of positive ion)A. `v^(-2)`B. `v^(-1)`C. `v`D. `v^(2)` |
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Answer» Correct Answer - C In cyclotron Force on charge = centripetal force `therefore " " Bqv = (mv^(2))/(r)` or `r = (mv)/(Bq)` `r prop v` |
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| 4. |
The angle `theta` between the vector `p=hati+hatj +hatk` and unit vector along X-axis isA. `cos^(-1) ((1)/(sqrt3))`B. `cos^(-1) ((1)/(sqrt2))`C. `cos^(-1) ((sqrt3)/(2))`D. `cos^(-1)((1)/(2))` |
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Answer» Correct Answer - A The angle between `p = hati + hatj + hatk ` `x -` axis ` x = hati` is given by `cos theta = (p * x)/(|p||x|) = ((hati + hatj + hatk))/(sqrt(1^(2) + 1^(2) + 1^(2)))((hati))/(sqrt(1^(2))) = (1)/(sqrt3)` `implies theta = cos^(-1) ((1)/(sqrt3))` |
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| 5. |
The massses of the three wires of copper are in the ratio 1 : 3 : 5. And their lengths are in th ratio 5 : 3 : 1. the ratio of their electrical resistance isA. `25 : 1 : 125`B. `1 : 125 : 25`C. `125 : 1 : 25`D. `125 : 25 : 1` |
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Answer» Given , Ratio of masses = `1 : 3 : 5` As density will remain same Ratio of volumes = `1 : 3 : 5 " " …. (i)` and Ratio of length `5 : 3 : 1 " " …. (ii)` For ratio of resistance `R_(1) : R_(2) : R_(3) = (rho l_(1))/(A_(1)) : (rho l_(2))/(A_(2) ) : (rho l_(3))/(A_(3))` Since , resistivity remain same ` = (l_(1))/(A_(1)) : (l_(2))/(A_(2)) : (l_(3))/(A_(3)) = (l_(1)^(2))/(A_(1) l_(1)) : (l_(2)^(2))/(A_(2) l_(2)) : (l_(3)^(2))/(A_(3) l_(3))` From Eqs . (i) and (ii) `R_(1) : R_(2) : R_(3) = ([5]^(2))/(1) : ([3]^(2))/(3) : ([1]^(2))/(5) = 25 : 3 : (1)/(5) = 125 : 15 : 1` |
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| 6. |
If average velocity of a sample of gas molecules at 300 K is `5 cm s^(-1)` , what is RMS velocity of same sample of gas molecules at the same temperature ? (Given , `alpha : u : v = 1 : 1 .224 : 1.127)`A. 6.112 cm/sB. 4.605 cm/sC. 4.085 cm/sD. 5.430 cm/s |
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Answer» Correct Answer - D Given that `alpha` (most probable velocity) u(RMS velocity) v ( average velocity) = `1 : 1.224 : 1 : 1.127` `therefore (u)/(v) = (1.124)/(1.127)` or `(u)/(5) = (1.224)/(1.127)` or `u = (1.124 xx 5)/(1.127) = 5.430` cm/s |
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| 7. |
A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude A. What will be the frequency of oscillation, the block will just start to slip? Coefficient of friction`=mu`.A. `(1)/(2pi ) sqrt((mu g)/(A))`B. `(1)/(4pi ) sqrt((mu g)/(A))`C. `2pi sqrt((A)/(mu g))`D. `4pi sqrt((A)/(mu g))` |
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Answer» Correct Answer - A When restoring force will become equal to the frictional force , block will start to slip . `therefore` Restoring force = Friction force `implies kA = mu mg " " … (i)` Now , frequency f `= (1)/(2pi) sqrt((k)/(m))` From Eq.(i) `f = (1)/(2pi) sqrt((mu g)/(A))` |
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| 8. |
In common base circuit of a transistor , current amplification factor is `0.95`. Calculate the emitter current , if base current is `0.2` mAA. 2 mAB. 4 m AC. 6 mAD. 8 mA |
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Answer» Correct Answer - B Given `alpha = 0.95 , I_(b) = 0.2` mA As `" " alpha = (I_(c))/(I_(e)) = (I_(e) - I_(b))/(I_(e)) = 1 - (I_(b))/(I_(e))` `implies (I_(b))/(I_(e)) = 1 - alpha = 1- 0.95` `implies (I_(b))/(I_(e)) = 0.5 implies (0.2)/(0.05) mA = I_(e)` `implies I_(e) = 4 mA` |
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| 9. |
Light of wavelength `lambda_(A)` and `lambda_(B)` falls on two identical metal plates A and B respectively . The maximum kinetic energy of photoelectrons in `K_(A)` and `K_(B)` respectively , then which one of the following relations is true ? `(lambda_(A) = 2lambda_(B))`A. `K_(A) lt (K_(B))/(2)`B. `2 K_(A) = K_(B)`C. `K_(A) = 2K_(B)`D. `K_(A) gt 2 K_(B)` |
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Answer» Correct Answer - A According to Einstein photoelectric equation `K_(A) = (hc)/(lambda_(A)) - phi implies 2K_(A) = (2hc)/(lambda_(A)) - 2 phi` or `2 K_(A) + phi = (2hc)/(lambda_(A)) - phi` Wavelength of metal B is half of metal A `therefore 2 K_(A) + phi = K_(B)` `implies K_(A) lt (K_(B))/(2)` |
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| 10. |
The relation between force `F` and density `d` is `F=(x)/(sqrt(d))`. The dimension of `x` isA. `[L^(-1//2) M^(3//2) T^(-2)]`B. `[L^(-1//2) M^(1//2) T^(-2)]`C. `[L^(-1) M^(3//2) T^(-2)]`D. `[L^(-1) M^(1//2) T^(-2)]` |
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Answer» Correct Answer - A Given `, F = (x)/(sqrtd)` Substituting dimensions `MLT^(2) = (x)/(sqrt(ML^(-3)))` `implies x = [M^(3//2) L^(-1//2) T^(2)]` |
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| 11. |
The velocity of water in river is 9 km/h of the upper surface . The river is 10 m deep . If the coefficient of viscosity of water is `10^(-2) ` poise then the shearing stress between horizontal layers of water isA. `0.25 xx 10^(-2) N// m^(2)`B. `0.25 xx 10^(-3) N//m^(2)`C. `0.5 xx 10^(-3) N//m^(2)`D. `0.75 xx 10^(-3) N//m^(2)` |
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Answer» Correct Answer - A Shearing stress = `(F)/(A)` and viscous force F = `eta A""(dv)/(dx)` `therefore ` Shearing stress = `(eta A"" (dv)/(dx))/(A) = eta "" (dv)/(dx)` `(10^(-2) "poise" [ 9 xx (5)/(18)] m//s)/(10 m) = 0.25 xx 10^(-2) N//m^(2)` |
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| 12. |
When a wave travels in a medium, the particle displacement is given by the equation `y=asin 2pi(bt-cx), ` where `a,b` and `c` are constants. The maximum particle velocity will be twice the wave velocity. IfA. b = acB. `b = (1)/(ac)`C. `c = pi a `D. `c = (1)/(pi a )` |
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Answer» Correct Answer - D Given y = a sin `2 pi ( bt - cx)` Comparing it will general equation `y = r sin [ (2 pi t)/(T) - (2pi)/(lambda) x]` We got , `(2pi )/(T) - omega = 2 pi b` and r = a `implies lambda = (1)/(c) ` and ` T = (1)/(b)` Maximum particle velocity `omega r = 2 pi ba ` Wave velocity ` v = (lambda)/(T) = (b)/(c)` Given maximum particle velocity = 2 `xx` wave velocity `2 pi ba = 2 xx (b)/(c) implies c = (1)/(pi a )` |
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| 13. |
In air , a charged soap bubble of radius r is in equilibrium having outside and inside pressures being equal . The charge on the drop is (`epsi_(0)` = permittivity of free space , T = surface tension of soap solution)A. `4 pi r^(2) sqrt((2 T epsi_(0))/(t))`B. `4 pi r^(2) sqrt((4 E epsi_(0))/(t))`C. `4 pi r^(2) sqrt((6 T epsi_(0))/(t))`D. `4 pi r^(2) sqrt((8 T epsi_(0))/(t))` |
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Answer» Correct Answer - D For a charged soap bubble `P_("external") + P_("electro") = (4T)/(r)` In equilibrium condition Surface charge density `sigma` is given by `sigma^(2) = (E epsi_(0)T)/(r) implies sigma = sqrt((8 epsi_(0) T)/(r))` `(q)/(4 pi r^(2)) = sqrt((8 epsi_(0)T)/(r)) implies theta = 4 pi rho^(2) sqrt((8 epsi_(0) T)/(rho))` |
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| 14. |
A sphere P of mass m moving with velocity u collides head on with another sphere Q of mass m which is at rest . The ratio of final velocity of Q to initial velocity of P is ( e = coefficient of restitution )A. `(e - 1)/(2)`B. `[(e+ 1)/(2)]^(1//2)`C. `(e + 1)/(2)`D. `[(e+1)/(2)]^(2)` |
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Answer» Correct Answer - C Here , `m_(1) = m_(2) = m , u_(1) = u,u_(2) = 0` Let `v_(1) , v_(2)` be their velocities after collision According to principle of conservation of linear momentum mu + 0 = `m(v_(1) + v_(2))` or `v_(1) + v_(2) = v " " … (i)` By definition , `e = (v_(2) - v_(1))/(u - 0) ` or `v_(2) - v_(1) = eu " " ... (ii)` Adding Eqs. (i) and (ii) , we get `v_(2) = (u (1 + e))/(2) implies (v_(2))/(u) = (1 + e)/(2)` |
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| 15. |
In electromagnetic wave , according to Maxwell , changing electric field givesA. stationary magnetic fieldB. conduction currentC. eddy currentD. displacement current |
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Answer» Correct Answer - D According to Maxwell time varying electric field produced displacement current . |
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| 16. |
Magnetic induction produced at the centre of a circular loop carrying current is B . The magnetic moment of the loop of radius R is (Me = permeability of tree space)A. `(BR^(3))/(2pi mu_(0))`B. `(2 pi BR^(3))/(mu_(0))`C. `(BR^(2))/(2 pi mu_(0))`D. `(2pi BR^(2))/(mu_(0))` |
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Answer» Correct Answer - B Magnetic moment `M = I xx A " " … (i)` Current `xx` Area of enclosed by loop Magnetic induction at the centre of circular loop `B = (mu_(0)I)/(2R) implies I = (2 BR)/(mu_(0))` Here , `A = pi R^(2) " " … (iii)` Substituting Eq.(ii)( and Eq.(iii) in Eq.(i) , we get `M = (2BR)/(mu_(0)) xx pi R^(2) = (2pi BR^(3))/(mu_0)` |
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| 17. |
A string of length L and force constant k is stretched to obtain extension l. It is further stretched to obtain extension `l_(1)`. The work done in second streching isA. `(1)/(2) kl_(1) ( 2l + l_(1))`B. `(1)/(2) Kl_(1)^(2)`C. `(1)/(2) K(l^(2) + l_(1)^(2))`D. `(1)/(2) K(l_(1)^(2) + l^(2))` |
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Answer» Correct Answer - D Work done in stretching a string to obtain an extension l is `W_(1) = (1)/(2) Kl` Similarly , work done is stretching a string to obtain extension `l_(1)` is `W_(2) = (1)/(2) kl_(1)^(2)` `therefore` Work done is second case `= W_(2) - W_(1) = (1)/(2) K(l_(1)^(2) - l^(2))` |
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| 18. |
The equiconvex lens has focal length f. If is cut perpendicular to the principal axis passin through optical centre, then focal length of each half isA. `(t)/(2)`B. tC. `(3t)/(2)`D. `2t` |
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Answer» Correct Answer - D When an equiconvex lens is cut parallel to principle axis focal length remains the same and when the lens is cut perpendicular to principle axis its focal length becomes twice the original . |
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| 19. |
Which among the following gases can be liquified easily ?A. ChlorineB. NitrogenC. OxygenD. Hydrogen |
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Answer» Correct Answer - A A gas can be liquified when its temperature is below critical temperature , by cooling or compressing it . Gases which have high critical temperatures (such as `Cl_(2) , NH_(3) , CO_(2) , SO_(2)` etc.) can be liquified by applying a suitable pressure alone . permanent gases (such as `H_(2) , N_(2) , O_(2) `etc.) cannot be liquified by the action of pressure and cooling . |
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| 20. |
20 mL solution of 0.1 M ferrous suplhate was completely oxidised using a suitable oxidising agent what is the number of electronic exchanged ?A. `1.204 xx 10^(22)`B. `193`C. `1930`D. `1.204 xx 10^(21)` |
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Answer» Correct Answer - D During oxidation of `FeSO_(4) , Fe^(2+)` ion is oxidises into `Fe^(3+)` ion. `Fe^(2+) to Fe^(3+) + e^(-)` i.e, `1e^(-)` is transferred from 1 molecule of `FeSO_(4)` , i.e, from 1 `Fe^(2+)` ion . Now , molarity of `FeSO_(4)` solution = 0.1 M and volume of solution = 20 mL = `(20)/(1000)` L `therefore` Number of moles = `0.1 xx (20)/(1000) = 0.002` `because` In 1 moles `FeSO_(4) ` , number of molecules = `6.022 xx 10^(23)` `therefore` In 0.002 mol `FeSO_(4)` , number of molecules `= 6.022 xx 10^(23) xx 0.002 = 1.204 xx 10^(21)` `therefore` Number of electrons transferred `= 1.204 xx 10^(21)` |
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| 21. |
Two concentric spheres kept in air have radii R and r. They have similar charge and equal surface charge density `sigma`. The electrical potential at their common centre is (where, `epsi_(0) =` permittivity of free space)A. `(sigma (R + r))/(epsi_(0))`B. `(sigma (R - r))/(epsi_(0))`C. `(sigma (R + r))/( 2 epsi_(0))`D. `(sigma ( R + r))/(4 epsi_(0))` |
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Answer» Correct Answer - A Potential at the centre due to the sphere 1 `V_(1) = (1)/(4pi epsi_(0)) (q)/(R)` and due to sphere `2 ` is `V_(2) = (1)/(4pi epsi_(0)) (q)/(r)` `therefore ` Electric potential at the common centre `V = V_(1) + V_(2) = (1)/(4pi epsi_(0)) (q)/(R) + (1)/(4 pi epsi_(0)) (q)/(r)` `= (q)/(4 pi epsi_(0)) [ (1)/(R) + (1)/(r)] = (q)/(4 pi epsi_(0)) [( R + r)/(Rr)]` We have `(q)/(4 pi R r) = sigma therefore v = (sigma (R + r))/(epsi_(0))` |
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| 22. |
Which statement is not correct about fullerene `C_(60)`?A. It contains 20 six membered rings and 12 five membered ringsB. All carbon atoms undergo `sp^(2)` hybridisationC. A six membered ring is fused with six membered rings onlyD. A five membered ring is fused with six membered ring only |
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Answer» Correct Answer - C `C_(80)` (fullerence) looks like a soccer ball and contains 20 six membered rings and 12 five membered rings of carbon atoms . Six membered rings are fused with both six membered as well as five membered rings while five membered rings are attached only with six membered rings . |
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| 23. |
What is the molality of solution containing 200 mg of urea `("molar mass"60" g mol"^(-1))` dissolved in 40 g of water ?A. `0.0825`B. `0.825`C. `0.498`D. `0.0013` |
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Answer» Correct Answer - A Molality (m) `= ("mass of solvent (in g)" xx 1000)/("molecular weight of solute" xx "mass of solvent (in g)")` `= (200 xx 10^(-3) xx 1000)/(60 xx 40) = 0.0833` |
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| 24. |
In which of the following oxides of nitrogen, the oxidation state of the element is the lowest ?A. Nitric oxideB. Nitrous oxideC. Nitrogen dioxideD. Nitrogen trioxide |
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Answer» Correct Answer - B (a) Nitric oxide , NO x - 2 `= 0 implies x = + 2` (b) Nitrous oxide ` , N_(2)O 2x - 2 = 0 implies x = + 1` (c) Nitrogen dioxide , `NO_(2)` x + 2(-2) = 0 `implies x = + 4` (d) Nitrogen trioxide `NO_(3)` `x + 2x (-3) = 0 implies x = + 6` Thus , in nitrous oxide , oxidation number of nitrogen is lowest . |
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