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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Convex lens has a focal length of 10cm a) where should the object be placed if the image is to be 30 cm from the lens on the same side as the object? (b) What will be the magnification? |
Answer» Solution : In case of magnifying lens, the lens is convergent and the image is erect, ENLARGED, virtual, between infinity and object and on the same side of lens as shown in figure. So here f=10 cm and v=-30cm HENCE from lens -formula. `1/v-1/u=1/f` we have `1/(-30)-1/u=1/10` i.e., u=-7.5 cm So the object must be PLACED in front of lens at a distance of 7.5 cm (which is `LT f`) from it. (b) `m=[I/O]=v/u=(-30)/(-7.5)=4` i.e., image is erect, virtual and four times the size of object. |
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| 2. |
In the following situation which one of the following is a correct statement ?When a body moves on circular path , the : |
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Answer» magnitude of velocity changes and its direction remain constant |
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| 3. |
A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for eachsection of the cycle. C_V=3/2R (a) A to B : constant volume (b) B to C: constant pressure (c) C to D: adiabatic (d) D to A: constant pressure |
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Answer» Solution :(a)For process A to B volume is CONSTANT hence WORK done dW=0 `rArr` From first LAW of thermodynamics , dQ=dU+dW `therefore` dQ=dU+0 `therefore` dQ=dU `=nC_V DT` `=nC_V(T_B-T_A)` `=(1)(3/2R)(T_B-T_A)` `=3/2(RT_B-RT_A)` `=3/2(P_B V_B -P_A V_A)` because in EQUATION of gas PV = RT at point A, `P_A V_A =RT_A` and at point B, `P_B V_B=RT_B` `therefore` Heat enchanged `=3/2 (P_B V_B-P_A V_A)` (b) For process B to C pressure is constant hence work done, `dW=PDeltaV` `therefore dW=P_B(V_C-V_B)` `rArr` From first law of thermodynamics, dQ=dU+dW `=3/2R(T_C-T_B)+P_B(V_C+V_B)` `therefore dQ=3/2 (RT_C -RT_B) +P_B (V_C -V_B)` `=3/2(P_C V_C- P_B V_B) + P_B (V_C-V_B)` `=5/2 P_B (V_C -V_B)` [ `because P_B=P_C` and `P_B=P_A` ] `therefore` Heat exchanged `=5/2P_B (V_C-V_A)` (d) For process C to D , it is adiabatic process so heat exchanged dQ=0 (d) For process D to A at constant pressure `P_A` the gas compressed from volume `V_D` to `V_A` The heat exchanged will be as shown in (b) `P_A` instead of `P_B`, `V_A` instead of `V_C` and `V_A` instead of `V_D` have to be taken. `therefore`Heat exchanged `dQ=5/2 P_A(V_A-V_D)` |
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| 4. |
Choose the correct statement A: It is impossible to derive continuoussupply of work by cooling a body to a temparature lower than that of the coldestof its surroundings. B: Heat engine can convert whole of the heatenergy supplied to it into useful work. |
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Answer» only A |
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| 5. |
A barometer kept in an elevator accelerating upward reads 76 cm. The air pressure in the elevator is |
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Answer» 76 cm |
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| 6. |
Two identical vessels A & B contain equal amount of ideal monoatomic gas. The piston of A is fixed but that of B is free. Same amount of heat is absorbe3d by A & B 's internal energy increases by 100 J the change in internal energy of A is |
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Answer» `100 J ` `triangleQ = TRIANGLEU for A` |
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| 7. |
If the earth, were suddenly contracted to (l/n)th of its present radius without change in its mass, then what will be the effect on the duration of the day. |
| Answer» SOLUTION :The duration will become (`24//n^2`) HOUR, because lay = a constant `(2/5) MR^2 xx (2pi//T)` = a CONST, `T prop R^(2)`when it contracts, `T^(1) prop (R^(2)//n^(2)) . T. =(T//n^(2))` or `T. = (24//n^(2))`. | |
| 8. |
In which method of heat transfer does gravity not play any part ? |
| Answer» SOLUTION :In CONDUCTION and RADIATION. | |
| 9. |
A body of mass m is kept over an object of mass M. The object is at rest on smooth. Coefficient of static friction between the two bodies is mu . What is theminimum force that needs to be applied on the object, so that the body will be about to slip over it? |
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Answer» Solution :Let the force applied onthe object horizontally Fig, so that the body is about to slide be F. Hence the common acceleration of the system, a `= (F)/(M+m),` So the force on the body = ma = `(Fm)/(M+m)` When the body is about to slip this force is balanced by the force of limiting FRICTION acting at the plane of contact. `:.` Frictional force , f = `MU MG = (Fm)/(m+M)` or, F`=mu (M+m)G`. 
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| 10. |
The driver of a truck travelling with a velocity v suddenly notices a wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into the wall? |
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Answer» Solution :To stop the truck by applying brakes, the final velocity should be ZERO. So, applying the EQUATION `v^(2) = u^(2) - 2 `aswe get, 0 = `v^(2) -2 ` ad or,a = `(v^(2))/(2D)` = retardation of the truck. If the mass of the truck is m , then the force applied by the brakes, F = ma = `(mv^(2))/(2d)` If the truck turns along a circular path to avoid collision, without applying brakes,then the radius of the that circular path = d. For this circular motion, the necessary centripetal force is, F.= `(mv^(2))/(d)` Here, F`lt`F. , hence, it is better to stop the truck applying brakes so that crashing CAB be avoided by the application of a comparatively less force. |
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| 11. |
Water in a river 20 m deep is flowing at a speed of 10ms^(-1). The shearing stress between the horizontal layers of water in the river in Nm is (coefficient of viscosity of water = 10^(3) SI units) |
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Answer» `2XX10^(-2)` |
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| 12. |
The surface tension phenomenon is the result of the tendency of a system to ……………. . |
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Answer» conserve energy |
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| 13. |
A cyclist is going round a circular track with a speed of 10 mcdot s^(-1). If the radius of the track is 80 m, calculate the angular speed of the cyclist. |
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Answer» |
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| 14. |
An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. IF the object is displaced through a distance of 20cm towards the mirror, the displacement of the image will be |
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Answer» 30CM AWAY from the mirror |
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| 15. |
When does two vectors equal magnitude have their resultant equal in magnitude of either of them? |
| Answer» SOLUTION :`R=sqrt(R^2+R^2+2R^2costheta),R^2+2R^2+2R^2costhetaR^2=2R^2(1+COSTHETA),1+costheta=1/2,costheta=(-1)/2,theta-120^@` | |
| 16. |
Four particles each of mass 100g are placed at the corners of a square of side 10 cm. Find the moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane. Find also the radius of gyration of the system. |
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Answer» Solution :MASS = m = 100 g, radius `r=5sqrt(2)CM` Moment of INERTIA `I=4mr^(2)` `= 4xx100xx(5sqrt(2))^(2)=2XX10^(-3)KGM^(2)` `k = sqrt((I)/(M))=sqrt((2xx10^(-3))/(0.4))=0.07 m = 7 cm`. |
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| 17. |
Derive the equation of angular momentum in the case of rotational motion about a fixed axis. |
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Answer» Solution :ANGULAR MOMENTUM of a particle of rigid body is in the direction of a fixed Z-axis `therefore vec(l_(1))=vec(r_(1))xxvec(p_(1))` If the rigid body consisting of n particle, then the vector sum of angular momentum of all particle will be obtain in the direction of Z-axis. `therefore vec(L_(Z))=vec(l_(1))+vec(l_(2))+vec(l_(3))+....vec(l_(n))` `=(vec(r_(1))xxvec(p_(1)))+(vec(r_(2))xxvec(p_(2)))+(vec(r_(3))xxvec(p_(3)))+....(vec(r_(n))xxvec(p_(n)))` `=underset(i=1)overset(n)sumvec(r_(i))xxvec(p_(i))"where "i=1,2,3,...,n` `=underset(i=1)overset(n)sumvec(r_(i))xxm_(i)vec(v)_(i)` `=underset(i=1)overset(n)summ_(i)r_(i)^(2)omegahatk"" [because v_(i)=r_(i)omega]` `=underset(i=1)overset(n)sumI_(i)omegahatk"where "underset(i=1)overset(n)summ_(i)r_(i)^(2)=I` but for all particle `SigmaI_(i)=I`, `vec(L_(Z))=Iomegahatk` where `hatk` is a UNIT vector in Z-axis. but angular momentum `vec(L_(1))` perpendicular to the axis in rigid body, then total angular momentum `vecL=vec(L_(Z))+vec(L_(_|_))` but `vec(L_(_|_))=0` `therefore vecL=vec(L_(Z))` `vecL=Iomegahatk` `therefore L=Iomega` SIMILAR to `p=mv` in linear motion. |
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| 18. |
Two parallel glass plates are dipped perpendicularly in a liquid of density rho. The separation between the plates is ‘d' and the surface tension T. The angle of contact of glass istheta . The capillary rise of the liquid between the plates is |
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Answer» ` (( T cos THETA) /( PD))` |
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| 19. |
A bullet of mass 2.5 g moving with a velocity of 500 ms^(-1) enters a wooden block and comes out of it with a velocity of 100 ms^(-1). Find the work done by the bullet while passing through the wooden block. |
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Answer» Solution :Mass `m = 2.5 g = 2.5 xx 10^(-3)` kg, Initial VELOCITY U = 500 m/s , Final velocity v = 100 m/s By WORK energy theorem, `W = (1)/(2) m u^(2) - (1)/(2) mv^(2) = (1)/(2) xx 2.5 xx 10^(-2) [(500)^(2) - (100)^(2)]` `= 1.25 xx 10^(-3) xx 10^(4) [25 - 1] therefore` Work done = 300 J |
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| 20. |
A sphere contracts in volume by 0.01 % when taken to the bottom of sea 1 km deep. Find the bulk modulus of the material of the sphere. |
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Answer» SOLUTION :Here `DELTA V = 0.01% ` of V ` = (0.01)/(100) xx V = 10^(-4) V` ` P = H RHO g= 10^3 xx 10^3 xx 9.8` ` = 9.8 xx 10^6 NM^(-2)` ` K = (PV)/(Delta V) = (9.8 xx 10^6 xx V)/(10^(-4) xx V) ` ` = 9.8 xx 10^10 Nm^(-2)` |
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| 21. |
A person moving horizontally with velocity barV_(m).Rain falls vertically with velocity barV_(R). To save himself from the rain, he should hold an umbrella with vertical at an angle of |
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Answer» (a) `tan^(-1)((V_(R))/(V_(m)))` |
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| 22. |
A neutron moving with a certain kinetic energy collidies head on with an atom of mass number A. The fractional kinetic energy retained by it is |
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Answer» `(A-1)/(A+1)` |
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| 23. |
For an ideal gas if the molar heat capacity varies as C = C_v + 3aT^2, find the equation of the process in the variables (T, V) where a is a constant. |
| Answer» SOLUTION :`Ve^(-((3A)/(2R))T^2)= ` CONSTANT | |
| 24. |
A rail track made of steel having length 10 m is clamped on a railway line at its two ends as shown in figure. On a summer day due to rise in temperature by 20^(@)C, it is deformed as shown in figure. Find x (displacement of the centre) if alpha_("steel")=1.2xx10^(-5)" "^(@)C^(-1) |
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Answer» SOLUTION :By using Pythagoras theorem, for given right angle triangle, `((L+DELTAL)/(2))^(2)=(L/2)^(2)+x^(2)` `x=SQRT(((L+DeltaL)/(2))^(2)-(L/2)^(2))` `=1/2(sqrt((L+DeltaL)^(2)-L^(2)))` `=1/2[sqrt(L^(2)+2LDeltaL+DeltaL^(2)-L^(2))]` `=1/2[sqrt(2LDeltaL+DeltaL^(2))]` As `DeltaL` is very small, `DeltaL^(2)` is neglected `x=1/2sqrt(2LDeltaL)` But `DeltaL=LalphaDeltaT` `:.x=1/2sqrt(2LxxLalphaDeltaT)` `=1/2Lsqrt(2alphaDeltaT)` `:.x=10/2xxsqrt(2xx1.2xx10^(-5)xx20)` `=5xxsqrt(4.8xx10^(-4))` `=5xx2xx1.1xx10^(-2)` `=11xx10^(-2)` `=0.11m=11cm` Here, the INCREASE in length of mercury is neglected as it is very small. |
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| 25. |
Following are three equations of motion S(g)=ut+(1)/(2)at^(2)v(s)=sqrt(u^(2)+2as)v(t)=u+at Where ,S,u,t,a,v are respectively the displacement ( dependent variable ), initial ( constant ), time taken ( independent variable ), acceleration ( constant ) and final velocity ( dependent variable ) of the particel after time t. Find the velocity of a particle after 100m- |
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Answer» `10 m//s` `v=0+2xx10` `=20m//s`. |
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| 26. |
An infinite number of particles each of mass m are placed on the positive x-axis at 1m, 2m, 4m, 8m, ----- from the origin. The magnitude of the resultant gravitational force on mass .m. kept at the origin is |
Answer» SOLUTION : The resultant gravitational force `F=(Gm^2)/1+(Gm^2)/4+(Gm^2)/16`+… `=Gm^2`(`1/1+1/4+1/16`+…..) = `Gm^2(1/(1-1/4))` (`because` sum of the infinite termsin G.P. is `S_prop =a/(1-r)`) `=4/3Gm^2` |
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| 27. |
Which of the following are axial vectors. |
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Answer» |
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| 28. |
A particle of mass m is acted upon by a force F given by the empirical law F + (R)/(t^2) v (t). If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot: |
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Answer» `logv (t) " against " 1/t` |
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| 29. |
A 1N pendulum bob is held at an angle thetafrom the vertical by a 2 N horizontal force F as shown in the figure. The tension in the string supporting the pendulum bob (in newton) is |
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Answer» `cos theta` |
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| 30. |
Three liquids A, B and C having same specific heat and masses m, 2m and 3m have temperatures 20^(@)C, 40^(@)C and 60^(@)C respectively. Temperature of the mixture when |{:("Column I","Column II"),("i. A and B are mixed at","(A) "35^(@)C),("ii. A and C are mixed at","(B) "52^(@)C),("iii. B and C are mixed at","(C) "50^(@)C),("iv. A, B and C all three are mixed at","(D) "45^(@)C),(,"(E)none"):}| |
| Answer» SOLUTION :i. (E), II. (A), iii. (D), iv. (B) | |
| 31. |
The radius of a sphere is (4.5+-0.2)cm. What is the percentage error in volume? |
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Answer» |
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| 32. |
Efficiency of a heat engine whose sink is at temperature of 300 K is 40% . To increase the efficiency to 60 %, Keeping the sink temperature constant, the source temperature must increased by |
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Answer» SOLUTION :`T_(2)/T_(1)=1-eta=1-40/100=3/5` `rArrT_(1)=5/3T_(2) rArrT_(1)=5/3xx300=500K` NEW EFFICIENCY `eta=60%` `T_(2)/T_(1)=1-eta^(.)=1-60/100=2/5` `T_(1)^(.)=5/2xx300=750K, DeltaT=750-500=250K` |
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| 33. |
Consider that an ideal gas (n moles) is expanding in a process given by p = f (V), which passes through a point (V_0, p_0). Show that the gas is absorbing heat at (p_0, V_0) if the slope of the curve p =f (V) is larger than the slope of the adiabatic passing through (p_0,V_0). |
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Answer» Solution :Slope of graph P=f(V) at POINT `(V_0, p_0)=f(V_0)` For adiabatic process `PV^(gamma)`=K (Constant) `therefore` At point `(V_0,P_0)P_0 V_0^gamma=k` `therefore P_0=k/V_0^gamma` `therefore dP_0=k(-gamma)V_0^(-gamma-1). dV_0` `therefore (dP_0)/(dV_0)=-gamma k/V_0^gamma xx 1/V_0` `therefore` Slope =`-gamma P_0/V_0`....(1)`[because k/V_0^gamma =P_0]` Now dQ=dU+dW `=nC_V dT + PdV`...(2) and PV =nRT `therefore T=(PV)/(nR)=(f(V)V)/(nR)` `therefore dT=1/(nR)[f(V)dV+Vf.(V)dV]` `=1/(nR) [f(V)+Vf.(V)]dV`...(3) `rArr` Equ. (2) `dQ=nC_V xx1/(nR)[f(V)+Vf.(V)]dV` `therefore ["dQ"/"dV"]_(V=V_0) =C_V/R[f(V_0)+V_0f.(V_0)]+f(V_0)` ...(4) Now, `C_P-C_V=R` `therefore C_P/C_V-1=R/C_V` `gamma-1=R/C_V` `therefore C_V/R=1/(gamma-1)`...(5) Putting value of (5) in (4), `((dQ)/(dV))_(V=V_0)=(C_V/R+1)f(V_0)+V_0f.(V_0)` `=[1/(gamma-1)+1]f(V_0) +(V_0f.(V_0))/(gamma-1)` `=gamma/(gamma-1)P_0+V_0/(gamma-1)f.(V_0) [ because f(V_0)=P_0]` When gas expended then `(dQ)/(dV) GT 0` and heat is absorbed. Now `(dQ)/(dV) gt 0` hence , `(gammaP_0)/(gamma-1)+V_0/(gamma-1) f.(V_0) gt0` `therefore gammaP_0+V_0f.(V_0) gt 0` `therefore f.(V_0)=-(gammaP_0)/V_0` |
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| 34. |
An ideal gas of specific heat ratio gammawith volume and pressure P is compressed to increase the pressure by DeltaP. If the change in volume is Delta V_1in an isothermal process and Delta V_2in an adiabatic process, then |
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Answer» `(DELTA V_1)/(Delta V_2) = GAMMA` |
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| 35. |
Load-elongation graphs of two wires A and B, made of the same material and of equal initial length are shown in the fig.1.21. Which wire is thicker? |
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Answer» Solution :`Y=(FL)/(al)` or, `a=(FL)/(YL)` (a= area of CROSS section of the wires) F,L and Y are the same for the TWO wires, `therefore a prop 1/l` So the wire of SMALLER elongation has a LARGER cross section. `therefore (a)_A gt (a)_B` Hence , the wire A is thicker than the wire B. |
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| 36. |
One annular metal ring of inner radis 7 cm and outer radius 14 cm and negligible weight is floating on the surface of a liquid. If surface tension of liquid is 0.08Nm^(-1), calculate the force required to detach it from liqudi surface. |
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Answer» SOLUTION :The contact length of annular ring with LIQUID surface `L=2pir_(1)+2pir_(2)L=2xx22/7xx(14+7)cm` `L=132cm=1.32m` Theh force required `F=S(L)=0.08xx1.32` `F=0.1056N` |
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| 37. |
Two small magnets of each of magnetic moment M_(0) is placed parallel to each other (show in figure). The magnetic field at point O is |
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Answer» ZERO `B_(2)=(mu_(0)M_(0))/(4pir^(3))` `therefore B=B_(1)-B_(2)=0` 
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| 38. |
Force acting on a particle is (2hat(i)+3hat(j)) N. Work done by this force is zero, when a particle is moved along the line 3y+kx=5. Here the value of k is. |
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Answer» 2 |
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| 39. |
A body cannot roll without slipping on a |
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Answer» ROUGH HORIZONTAL surface |
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| 40. |
An insect trapped in a circular groove of radius 12 cm moves along the groov steadily and completes 7 revolutions in 100 s . (a)What is the angular speed , and the linear speed of the motion ? (b)Is the acceleration vector a constant vector ? What isits magnitude ? |
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Answer» Solution :This is a example of uniform circular MOTION . HereR= 12 cm . The ANGULAR speed `omega` is given by `omega=2pi//T=2pixx7//100=0.44 " red "//s` The linear speed V is : `v=omegaR0.44s^(-1)xx12cm=5.3cms^(-1)` The direction of velocity v is along the tangent to the circle at EVERY point . The acceleration is directedtowards the centre of the circle . Since this direction changes continuously, acceleration here is not a CONSTANT vector . However , the magnitude of acceleration is constant : `a=omega^(2)R=(0.44s^(-1))^(2)(12cm)` `=2.3 " cm "s^(-2)` |
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| 41. |
A body is projected at angle 30^(0) to horizontal with a velocity 50 ms^(-1). Its time of flight is (g = 10 m//s^(2)) |
| Answer» Answer :B | |
| 42. |
Gas al a pressure P_0 is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to |
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Answer» <P>`4P_(0)` |
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| 43. |
Deduce the relation between angular momentum and angular velocity. |
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Answer» Solution :Let US consider the point mass m is at a distance r from the axis of rotation. Its linear momentum at any instant is tangential to the circular path. Then the angular momentum `vecL` is perpendicular to `vecr` and `vecp`. Hence it is directed along the axis of rotation. The angle `theta` between `vecr` and `vecp` in this case is `90^(@)`. The magnitude of the angular momentum L could be written as `LK=r mv sin 90^(@)=r mv` where v is the linear velocity. The relation between linear velocity v and angular velocity `OMEGA` in a circular MOTION is `v=r omega`. Hence `L=r m r omega` `L=(mr^(2))omega` THe directions OFL and `omega` are along the axis of rotation. the abvoe EXPRESSION can be written in the vector notation as `vec=L(=mr^(2))vec(omega)` `vec(L)=(Sigma m_(i)r_(i)^(2)vec(omega)` `vecL=Ivec(omega)` |
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| 44. |
The electric force between two electric charges is given by F=k(q_(1)q_(2))/(r^(2)) where r is the distance between q_(1) and q_(2) . Give the dimensional formula of k. |
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Answer» `M^(1)L^(2)T^(4)A^(2)` |
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| 45. |
The centripetal force on a particlemoving in a uniform circular motion on a horizontal circle of radius r is(-sigma)/(r^(2)) , thenfind themechanical energy of this particle . |
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Answer» Solution :If the centripetal FORCE `F_(C)` is in the uniform circular motion , then ` | F_(C )| == SIGMA/(r^(2))` ` :. (mv^(2))/r = (sigma)/(r^(2))` ` :. Mv^(2) = siga/r` ` :. ` Kinetic energy `1/2 mv^(2) = sigma/(2r) ""…(1)` and for potential energy , `F = - (dV)/(dr)` ` :. dV = 0 Fdr` ` :. dV = -Fdr` ` :. int dV = int - Fdr ""( :. F = F_(C))` ` :. V = int -(-sigma/(r^(2)))dr` ` = sigma int 1/(r^(2)) dr` ` = sigma [-1/r]` ` :. V = -sigma/r "" .....(2)` MECHANICAL energy `E = K+V` ` = sigma/(2r) -sigma/r [ :. " From equation (1) and (2) "] ` ` :. E = - sigma/(2r)` |
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| 46. |
A billiard ball of mass m and radius 7, when hit in a horizontal direction by a cue at a height h above its centre, acquired a linear velocity v_(0). The angular velocity Wo acquired by the ball is |
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Answer» `(5v_(0) r^(2))/(2H)` As is clear from figure  Angular momentum = MOMENT of momentum `I omega_(0)= (mv_(0) ) h or (2)/(5) mr^(2) omega_(0) = mv_(0) h or omega_(0) = (5 v_(0) h)/(2r^(2))` |
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| 47. |
Abody of mass m is suspended from two light springs of force constant k_(1) and k_(2)(k_(1)) separately. The periods of vertical oscillations are T_(1) and T_(2) respectively. Now the same body is suspended from the same two springs which are first connected in series and then in parallel. The period of vertical oscillations are T_(s) and T_(p) respectively, then |
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Answer» `T_(p)ltT_(1)ltT_(2)ltT_(s)` `T_(s)=2pisqrt((m)/((k_(1)k_(2))//(k_(1)=k_(2))))` `T_(p)=2pisqrt((m)/((k_(1)+k_(2))))` From above we note that `T_(s)gtT_(2)gtt_(1)gtT_(p),i.e.,` OPTION (a) is CORRECT. `(1)/(T_(p)^(2))=(k_(1)+k_(2))/(4pi^(2)m)=(k_(1))/(4pi^(2)m)+(k_(2))/(4pi^(2)m)` `=(1)/(T_(1)^(2))+(1)/(T_(2)^(2)),i.e.,` option (d) is correct. `T_(s)^(2)=(4pi^(2)m(k_(1)+k_(2)))/(k_(1)k_(2))=(4pi^(2)m)/(k_(2))+(4pi^(2)m)/(k_(1))` `=T_(2)^(2)+T_(1)^(2), i.e.,` option (b) is correct. |
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| 48. |
When…………..acts on a system of particles, then………….of the system remains…………… . |
| Answer» Solution :no EXTERNAL torque , angular MOMENTUM , a CONSTANT. | |
| 49. |
Give the expression for the speed of the transverse wave on a stretched string along with the meanings of the symbols used. |
| Answer» Solution :The VELOCITY of the wave on a string `V = SQRT((T)/( mu ))`where 'T' is TENSION in the string and `'mu'` is linear mass density of the MATERIAL of the string. | |
| 50. |
The pressure of the Earth's atmosphere at sea level is due to the ………….. . |
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Answer» GRAVITATIONAL attraction of the Earth for the atmosphere |
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