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.
| 37601. |
Area enclosed by x rarr t graph gives distance covered. |
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| 37602. |
At what temperature, hydrogen molecules will escape from the earth's surface? (Take mass of hydrogen molecule =0.34 xx 10^(-6) kg, Boltzman constant =1.38 xx 10^(-23) KJ^(-1). Radius of earth =6.4 xx 10^(6)m" and acceleration due to gravity "=9.8 ms^(-2)) |
| Answer» Answer :D | |
| 37603. |
A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius 100/2pi cm at a constant angular speed of 5-0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air = 332 m s^-1. |
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Answer» `Radius of the circle `=100/picm` `=100/pixx10^-2m` `=(1/pi)METRES` `omega=5rev/sec` So the linear speed `v=omegar=5/pi=(1.59)` so, velocity of the SOURCE `V_s=1.59m/s` so the at the position A the observer will listen maximum and at the position B it will listen MINIMUM frequency. So apparent frequency at A `=(332/(332+159))xx500~~515Hz` So apparent frequency at B `=(332/((332+159))xx500~~485GHz` |
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| 37604. |
If the time period (t) of vibration of a liquid drop depends on density (rho) of the liquid, radius (r) of the drop and surface tension (S), then the expression of t is where k is a dimensionless constant. |
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Answer» `t= ksqrt((rhor^(3))/(S))` where k is a dimensionless constant Equating dimensions on both sides, we get `[T]= [ML^(-3)]^(x)[L]^(y)[ML^(0)T^(-2)]^(z)` or `[M^(0)L^(0)T]= [M^(x+z)L^(-3x+z)T^(-2Z)]` According to PRINCIPLE of homogeneity of dimensions we get x+z= 0...(i) -3x+y=0 ..(ii) -2z = 1 ...(iii) SOLVING equations (i), (ii), and (iii), we get `x= (1)/(2), y= (3)/(2), z= (-1)/(2) :. t= krho^(1//2)r^(3//2)S^(-1//2)` or `t= ksqrt((rhor^(3))/(S))` |
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| 37605. |
Two holes, each of area A = 0.2 cm^(2) are drilled in the wall of a vessel filled with water. The distances of the holes from the level of water are h and h + H. Find the point where the streams flowing out of the holes intersect. The level of water is maintained in the vessel by regulated supply. |
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| 37606. |
Find the angle .theta. with the lower vertical at which the resultant acceleration of the bob is along the horizontal, when the perpendicular is released from horizontal position. |
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Answer» <P> SOLUTION :Since the resultant acceleration is along the horizontal, the VERTICAL acceeleration at .P. should be zero.ie., `a_(c) sin(90 - theta) = a_(t) sin theta` `(a_(c))/(a_(1)) = tan theta ..... (1)` But `a_(c) = (v^(2))/(l)(2glcostheta)/(l)=2gcostheta` and `a_(1) = g sin theta` Then `(a_(c))/(a_(t)) = 2 cot theta` SUBSTITUTING in EQUATION (1) `2 cot theta = Tan theta` Hence `theta = tan^(-1) (sqrt(2))`
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| 37607. |
A bullet is fired normally on an immovable wooden plank of thickness 2 m. It loses 20% of its kinetic energy in penetrating a thickness 0.2 m of the plank. The distance penetrated by the bullet inside the wooden plank is |
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Answer» 0.2m |
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| 37608. |
The volume of a liquid (v) flowing per second through a cylindrical tube depends upon the pressure gradient (p/l) radius of the tube (r ) coefficient of viscosity (eta) of the liquid by dimensional method the correct formula is |
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Answer» <P>`V ALPHA (PR^(4))/(ETA l)` |
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| 37609. |
Explain in detail the idea of weightlessness using lift as an example. |
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Answer» Solution :When a man is standing in the elevator, there are two forces acting on him. 1. Gravitational force which acts downward. If we take the vertical direction as POSITIVE y direction, the gravitational force acting on the man is `vecF_G =- MG hatj` The normal force exerted by floor.on the man which acts vertically upward, `vecN = Nhatj`Weightlessness of freely falling bodies: Freely falling objects EXPERIENCE only gravitational force. As they fall freely, they are not in CONTACT with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is EQUAL to the acceleration due to the gravity of the i.e., (a = g) . Newton.s 2nd law acting on the man `N= mg (g - a) [ because a = g ] N= 0` |
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| 37610. |
A spring of force constant 'k' is stretched by a small length 'x'. The work done is stretching it further by a small length 'y' is |
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Answer» `(1)/(2)k(X^(2)+y^(2))` |
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| 37611. |
A simple pendulum with a metal bob has a time period 'T'. Now the bob is immersed in a liquid which is non viscous. Thus time period is '3T'. Then the ratio of densities of metal bob and that of liquid is |
| Answer» ANSWER :B | |
| 37612. |
In above problemabout an axisperpendicularto the palneof frameand passing through a cornerof framethe moment of inertiaof threebodies is |
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Answer» `ML^(2)` |
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| 37613. |
In the following question statement 1 represent Assertion and statement 2 represents Reason .Statement 1 : In a uniform circularmotion the momentum of a particle does not change with time .Statement 2 : The kinetic energy of the particle remains constnat .Choose the correct statement of the following statements |
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Answer» Statement 1 is true and statement 2 is true and it EXPLAINS statement 1. But its velocity remains constant . So kinetic energy remainsconstant . |
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| 37614. |
A body cools from 70^(@)C to 50^(@)C in 5 minutes. Temperature of surroundings is 20^(@)C. Its temperature after next 10 minutes is |
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Answer» `25^(0)C` |
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| 37615. |
{:("List -I","List -II"),((a)"fraction of total energy that is utilised to increase internal energy of a monoatomic gas ",(e)0.25),((b)"fraction of total energy that is involved in expansion for a triatomic gas ",(f) "infinity"),((c)"specific heat during isothermal process ",(g) zero),((d)"specific heat during adiabatic process",(h)0.6):} |
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Answer» a-h, b-e,C-g,d-h |
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| 37616. |
A horizontal disc is freely rotating about a vertical axis passing through its centre at the rate of 100 rpm. A bob of wax of mass 20 g falls on the disc and sticks to it a distance of 5 cm from the axis. If the moment of inertia of the disc about the given axis is 2xx10^(-4)kgm^(2), find new frequency of rotation of the disc. |
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Answer» SOLUTION :`I_(1)` Momnet of inertia of the disc `=2xx10^(-4)kgm^(2)` `I_(2)` = MOMENT of inertia of the disc+Moment of inertia of the bob of wax on the dise `=2xx10^(-4)+mr^(2)=2xx10^(-4)+20xx10^(-3)(0.05)^(2)` `=2xx10^(-4)+0.5xx10^(-4)=2.5xx10^(-4)kgm^(2)` `(n//t)_(1)=100" rpm" ,(n//t_(2))=?` By the principle of conservation of angular momentum, `I_(1)omega_(1)=I_(2)omega_(2)""I_(1)2pi((n)/(t))_(1)=I_(2)2pi((n)/(t))_(2)` `2xx10^(-4)xx100=2.5xx10^(-4)((n)/(t))_(2)` `((n)/(t))_(2)=(100xx2)/(2.5)=80" rpm"` |
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| 37617. |
A particle moves along a straight line retardation a = ksqrt(v) [ k is a positive constant ]. The velocity of the particle at time t = 0 is u. How far would the particle move before coming to rest ? |
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| 37618. |
(A): A wire may be stiffer than another wire B. But B may stronger than A (R): A high Young's modulus does not necessarily imply a high value for the breaking stress |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 37619. |
The coefficient of linear expansion alpha of a rod length 2m varies with distance x from one end as alpha=a+bx where a=1.5xx10^(-5)//""^(0)C and b=2.5xx10^(-6)//m-""^(0)C. Find the increase in the length of the rod when heated through 350^(0)C. |
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| 37620. |
A small hole is made at the bottom of a hollow sphere. The water enters into it when it is taken to a depth of 40 cm under water. If the surface tension of water is 0.07 Nm^(-1), diameter of the hole is |
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Answer» `10//7mm` |
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| 37621. |
The displacement r of a particle varies with time as x=4t^(2)-15t+25Find the position, velocity and acceleration of the particle at t = 0 |
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Answer» Solution :Position, x=25 Velocity `=(dx)/(dt)=8t-15` `t=0,v=0-15=-15m//s` ACCELERATION `a=(DV)/(dt)=8ms^(-2)`. |
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| 37622. |
Consider the system of weights P_(1)=1 kg,P_(2) = 2 kg,P_(3) = 5kg and P_(4) = 0.5 kg as shown in Figure 7.14. alpha = 30^(@), g = 9.8 ms^(-2) and the coefficient of friction between the weights and the planes is 0.2 Find (a) the acceleration of the system (b) tensions in the strings between P_(1) and P_(2) and that between P_(2) and P_(2) respectively and (c) the force with which P_(4) presses down on P_(3) [Hint : Make free-body diagrams and apply Newton's law of motion after resolving the forces acting on a body along and perpendicular to the plane on which it lies] |
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| 37623. |
A boat moves relative to water with a velcity which is .n. times the river flow (a) If n lt t boat can not cross the river (b) If n = 1 boat can not cross the river without drifting (c) Ifgt 1 boat ca cross the river along shortest path (d) Boat can cross the river what ever is the value of n excluding zero value |
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Answer» only a is correct |
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| 37624. |
A large tank of water at a constant temperature of 90^(0)C is connected to a small vessel containing 10kg water at 10^(0)C, through a metal rod of length 2m, area of cross-section 0.5m^(2) and thermal conductivity of 20 units. If the time taken for the temperature of the water in small vessel at 30^(0)C is t=Nxx4200log(4//3), then find the value of N. (Given, specific heat of water =4200" J kg"^(-1)""^(0)C^(-1)) |
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| 37625. |
Examine table.s and express a) The energy required to break one bond in DNA in eV, b) the kinetic energy of an air molecule (10^(-2) J in eV, c) The daily intake of a human adult in kilocalories. |
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Answer» Solution :a) Energy required to break one bond of DNA is: `(10^(-20)/(1.6 xx 10^(-19) J//eV) =~ 0.06 eV` Note, 0.1eV = 100meV (100 millielectron VOLT). B) the kinetic energy of an air molecule is `(10^(-21))/(1.6 xx 10^(-19)J//.eV=~ 0.0062eV` this is the same as 6.2 meV. c) The average human consumption in a DAY is `10^(7)/(4.2xx 10^(3)J/kcal) -~2400 kcal` We point out a common misconception created by newspapers and magzines. They mention food VALUES in calories and urge us to restrict diet INTAKE to below 2400 claories. What they should be saying in kilocalories (kcal) and not calories. A person CONSUMING 2400 calories a day will soon starve to death 1 food calorie is 1 kcal. |
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| 37626. |
Two bodies of masses m_1 and m_2are separated by certain distance. If vecF_(12)is the force on m_1due to m_2and vecF_(21)is the force on m_2due to m_1then |
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Answer» `F_(12) = F_(21)` |
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| 37627. |
A cyclist is riding with a speed of 27kmh^(-1). As he approaches a circular turn on the road of radius 80m, he applies brakes and reduces his speed at the constant rate of 0.50ms^(-1) every second. The net acceleration of the cyclist on the circular turn is |
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Answer» `0.68 MS^(-2)` |
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| 37628. |
Why does a child feel more pain when she falls down on a hard cement floor, than when she falls on the soft muddy ground in the garden ? |
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Answer» SOLUTION :WHENA childfallsona cementfloorthemomentumis reducedto zeroin comparativelylesserttime .DUE to thisrateof changeofmometumis largeso greaterforceact on himandchildfeels morepain. When childfalls on softmuddygroundmomentumis REDUCED incomparativelylargertimehenceforcedecreaseandchildfeel lesspain. |
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| 37629. |
The equation of motion of a particle is x=acos(alphat)^(2). The motion is |
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Answer» PERIODIC but not oscillatory as the equation of motionn INVOLVES COSINE function and `t^(2)` HENCE it is oscillatory but not periodic. |
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| 37630. |
A body is released from the top of a smooth inclined plane of inclination theta. It reaches the bottom with velocity v. If the angle of inclination is doubled for the same length of the plane, what will be the velocity of the body on reaching the ground |
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Answer» v |
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| 37631. |
Two identical metal rods A and B are joined end to end. The free end of A is kept at 27°C and the free end of B at 37°C. Calculate the temperature of the interface. Thermal conductivity of A = 385 "Wm"^(-1) "K"^(-1), that of B 110 "Wm"^(-1) "K"^(-1). |
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| 37632. |
According to parallelogram law of vector addition, the resultant is given by : |
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Answer» `SQRT(P^(2) +Q^(2) +2PQcos theta)` |
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| 37633. |
A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base .If the redius of the vessel is r and angular velocity of rotation is omega, then the difference in the heights of the liquid at the centre of the vessel and the edge is ..... |
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Answer» `(romega)/(2g)` `P_(A)+(1)/(2)rhov_(A)^(2)+rhogh_(A)` `=P_(B)+(1)/(2)rhov_(B)^(2)+rhogh_(B)` Here , `h_(A)=h_(B)` `thereforeP_(A)+(1)/(2)rhov_(A)^(2)` `=P_(B)+(1)/(2)rhov_(B)^(2)` `thereforeP_(A)-P_(B)=(1)/(2)rho[v_(B)^(2)-v_(A)^(2)]` Now ,`v_(A)=0,v_(B)-romegaandP_(A)-P_(B)=rhog` `thereforehrhog=(1)/(2)RHOR^(2)omega^(2)orh=(r^(2)omega^(2))/(2g)` |
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| 37634. |
Two bodies of masses m_(1) and m_(2) are moving at uniform speed along circular paths of radii r_(1) and r_(2) respectively. If they take equal time to describe the cricles completely, the ratio of their angular velocities will be |
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Answer» `(r_(1))/(r_(2))` |
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| 37635. |
A cylinderical vessel filled with water is released on a fixed inclined surface of angle theta as shown in figure. The friction coefficient of surface with vessel in mu(lt tan theta). Then the constant with the incline will be (Neglect the viscosity of liquid) |
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Answer» `TAN^(-1) MU` |
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| 37636. |
Fill in the blanks using the words from the list appended with each statement Viscosity of glass…….with temperature,whereas viscosity of liquids…..with temperature (increases/decreases) |
| Answer» SOLUTION :`eta` of GASES increases, `eta` of LIQUID DECREASES with temperature | |
| 37637. |
The area of the surface of water in a vessel is 40cm^2. It is transferred into a cylindrical vessel of area of cross section 220cm^2. If the surface tension of water is 0.07 Nm^(-1), the workdone in the transfer of water is neglect gravity. |
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Answer» 12.6 erg |
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| 37638. |
Consider a cycle tyre bering filled with air by a pump. Let V be the volume of the tyre (fixed) and at each stroke of the pump DeltaV(=V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P_(1)" to "P_(2) ? |
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Answer» Solution :`P(V+Deltaupsilon)^(GAMMA)=(P+DELTAP)V^(gamma)` `P[1+gamma(Deltaupsilon)/V]=P[1+(Deltap)/P]` `gamma(Deltaupsilon)/V=(Deltap)/P,(dupsilon)/(DP)=V/(gammaP)` W.D =`int_(P_(1))^(P_(2))Pdupsilon= int_(P_(1))^(P_(2))PV/(gammaP)dp=((P_(2)-P_(1)))/gammaV` |
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| 37639. |
The work done in moving a body of mass 5kg from the bottom of a smooth inclined plane to the top is 50 j. If the angle of inclination of the inclined plane is 30^(@), its length is (g = 10 ms^(-2)) |
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Answer» 1/2m |
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| 37640. |
(I) Net Torque produces linear motion in rigid object. (II) Rolling motion is the combination of translational and rotational motions. Which statement is Incorrect ? |
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Answer» I only |
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| 37641. |
What is rolling friction ? Write laws of rolling friction. Define coefficient of rolling friction. |
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Answer» Solution :Whenbody isrollingwithoutslidingfrictionforceis calledrollingfrictionit is denoted by`f_(r )` Duringrollingthe surfacein contactgetmomentarilydeformeda littleand smallpart ofbodyremainin contactwith surface Asa resultthere is componentof the contactforceparallelto surfacewhichopposesmotion `mu_(F ) ltmu_(k)GT mu_(s)` |
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| 37642. |
Which of the following quantity is dimensionless ? |
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Answer» ANGLE |
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| 37643. |
A system consists of three particles located at the corners of a right triangle as shown in the figure. Find the position vector of centre of mass of the system. |
Answer» Solution : Using the EQUATION `X_(c)=(summ_(i)x_(i))/(M)=(2md+m(b+d)+3m(d+b))/(6M)=d+(2//3)b` `Y_(c)=(summ_(i)y_(i))/(M)=(2m(o)+m(o)+3MH)/(6m)=H//2` `Z_(c)=0`, because the particles are in X - Y plane we can express the position of centre of mass from the origin using a position vector as `vecr_(c)=X_(c)hati+Y_(c)hatj+Z_(c)hatk,vecr_(c)=(d+(2)/(3)b)hati+(h)/(2)hatj` |
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| 37644. |
In a jar having a mixture of H_(2), and He |
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Answer» HYDROGEN has more MEAN KE |
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| 37645. |
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm. given that the surface tension of soap solution at the temperature (20^@C) is 2.50 times 10^-2 N m^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container the soap solution (of relative density 1.20] what would be the pressure inside the bubble?[1 atmospheric pressure is 1.01 times 10^5 Pa). |
| Answer» Solution :Excess pressure inside the soap BUBBLE =20.0 Pa excess pressure inside the AIR bubble in soap solution =10.0Pa OUTSIDE pressure for air bubble =`1.01 times 10^5+0.4 times 10^3 times 9.8 times 1.2=1.06 times 10^5 Pa`. The excess pressure is so small that up to THREE significant FIGURES, total pressures inside the air bubble is `1.06 times 10^5 Pa` | |
| 37646. |
A Carnot engine of efficiency 40% , takes heat from a source maintained at a temperature of 500 K . If is desired to have an engine of efficiency 60%. Then , the source temperature for the same sink temperature must be |
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Answer» Solution :`eta=1-(T_(2))/(T_(1)),0.4=1-(T_(2))/(500)RARR T_(2)=300K` `0.6=1-(T_(2))/(T_(1)^(1))=1-(300)/(T_(1)^(1))rArr T_(1)^(1)=(300)/(0.4)=750K` |
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| 37647. |
A body is rotating uniformly about vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is |
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Answer» vertical |
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| 37648. |
The torque acting on a planet due to gravitational force of attraction is |
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Answer» `PROP` |
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| 37649. |
The density of a non-uniform rod of length 1m is given by rho (x)=a (1+bx^(2)) where, a and b are constant and 0 le x le 1. The centre of mass of the rod will be at |
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Answer» `(3(2+B))/(4(3b+b))` where a and b are constants and `0 le x le 1` LET `b rarr 0`, in this case `rho (x) = a=` constant Hence, centre of mass will be at `x=0.5` m (MIDDLE of the rod) Putting, `b=0` in all the options, only (a) given 0.5. Note : We should not check options by putting `a=0`, because `rho = 0` for a = 0. |
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| 37650. |
A motorist drives north for 35.0 minutes at 85.0 km/h and then stops for 15.0 minutes. He next continues north, travelling 130 km in 2.00 hours. What is his total displacement |
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Answer» 85km |
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