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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.
| 12301. |
Consider an object of mass 2 kg resting on the floor. The coefficient of static friction between the object and the floor is mu_8= 0.8. What force must be applied on the object to move it? |
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Answer» Solution :SINCE the object is at REST, the gravitational force experienced by an object is balanced by normal force exerted by FLOOR. N = mg The maximum static frictional force `f_(s)^(max) = mu_sN = mu_s mg ` `f_(s)^(max) = 0.8 xx 2 xx 9.8 = 15.68 N` Therefore to move the object the external force should be greater than maximum static FRICTION. ` F_(ext) gt 15.68 N` |
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| 12302. |
Three particles, each of mass m are placed at the vertices of an equilateral triangle of side a. What are the gravitation field and gravitational potential at the centroid of the triangle. |
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Answer» Solution :Refer to Fig. `O` is the centriod of triangle `ABC`, where `OA = (2)/(3) AD = (2)/(3) (AB sin 60^(@))` `= (2)/(3) xx a xx (sqrt(3))/(2) = (a)/(sqrt(3))` Thus, `OA = OB = OC = (a)/(sqrt(3))`  The gravitational intensity at `O` DUE to mass `m` at `A` is, `I_(A) = (Gm)/((OA)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OA`. SIMILARLY the gravitational intensity at `O` due to mass `m` at `B` is, `I_(B) = (Gm)/((OB)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OB`. and gravitational intensity at `O` due to mass `m` at `C` is, `I_(C) = (Gm)/((OC)^(2)) = (Gm)/((a//sqrt(3))^(2))` along `OC`. As `I_(A),I_(B)` and `I_(C)` are EQUAL in magnitude and equally inclined to each other, the resultant gravitational intensity at `O` is zero. gravitational potential at `O` due to masses at `A,B` and `C` is `V = - (Gm)/(OA) + (-(Gm)/(OB)) + (-(Gm)/(OC))` `= - (3 Gm)/(OA) = (-3 Gm)/(a//sqrt(3)) = (-3 sqrt(3) Gm)/(a)` |
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| 12303. |
A container contains 35 kg water and 0.2 kg water leaks from the container. Find the amount of water in container. |
| Answer» Solution :35 kg | |
| 12304. |
The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin ((2 pi)/(3) x) cos (120 pi t) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 xx 10^(-2) kg. Answer the following : (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency , and speed of each wave ? (c ) Determine the tension in the string. |
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Answer» Solution :(a) STATIONARY WAVE (B) `l=3, n=60Hz, and v= 180 ms^(-1)` for each wave (C ) 648N |
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| 12305. |
One end of a copper rod of length 0.25m and area of cross-section 10^(-4) m^(2) is immersed in a liquid boiling at 125^(@)C whereas the other end is kept cold by dipping It in an ice-cold bath. Find (i) temperature gradient, (ii) the rate of heat transfer (iii) the temperature at a point 0.1m from the higher temperature end. (lamda of copper =92cals^(-1)m^(-1)K^(-1)) |
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| 12306. |
Two particles are thrown from the same point with the same velocity of 49 ms^(-1). First is projected making angle theta with the horizontal and second at an angle (90^(@)-theta). The second particle is found to rise 22.5 m higher than the first. Find the height to which each particle will rise. |
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| 12307. |
A small mass m is moved slowly from the surface of the earth to a height h above the surface. The work done (by an external agent) in doing this is (a) mgh, for all values of h (b)mgh , for h lt lt R (c) 1/2mgR, for h=R (d)-1/2mgR, for h=R |
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Answer» only a & B are true |
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| 12308. |
If vec(C ) = vec(A) - vec(B), then: |
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Answer» `bar(A) xx bar(C ) = bar(B) xx bar(C )` |
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| 12309. |
A particle performing SHM with a frequency of 5 Hz and amplitude 2cm is initially in the positive extreme position. What is the equation for its displacement? |
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Answer» `x=0.02cos10pit` |
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| 12310. |
A whistle emitting a sound of frequency 440 Hz is tied to a string of 1.5 m length and ratated with an angular velocity of 20 rad .s^(-1)in the horizontalplane . Find the range of frequencies heard by an observer stationed at a large distance from the whistle . Speed of sound V = 350 m . s^(-1), |
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| 12311. |
To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. |
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Answer» <P> Solution :Here `a=200" RADS"^(-1)`, TORQUE `tau=180Nm`Power P = torque `(tau)xx` angular speed `(OMEGA)` `=180xx200=36000` Watt = 36 kW |
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| 12312. |
A force of 5N is applied on a 20 kg mass at rest. The work done in the third second is, |
| Answer» Answer :A | |
| 12313. |
The value of 124.2 + 52.487 with due regard to significant places is |
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Answer» 176.7 |
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| 12314. |
__________ is an example for motion in 3 dimension |
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Answer» A BODY falling freely under GRAVITY close to earth |
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| 12315. |
A horizontal pipe of non - uniform cross - sectional allows water to flow through it with a velocity 1ms^(-1) when pressure is 50 kPa at a point. If the velocity of flow has to be 2ms^(-1) at some other point, the pressure at that point should be |
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Answer» 51.5 KPA |
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| 12316. |
The gravitational potential energy of a system of eight identical masses, each of mass m, places at the corners of a cube of side L is (Gm^(2))/(L) (12 + Nsqrt(2) + (4)/(sqrt(3))). Find the value of n |
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| 12317. |
If the road is horizontal then the normal force and gravitational force are |
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Answer» equal and ALONG the same DIRECTION |
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| 12318. |
The value of 124.2-52.487 with due regard to significant places is |
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Answer» 71.7 |
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| 12319. |
The displacement of a particle executing periodic motion is given by y= 4cos^(2)((t)/(2))sin(1000t). Find independent constituent simple harmonic motions. |
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Answer» Solution :`y= 4COS^(2)((t)/(2))sin(1000t)` `y= 2[1+ cost]sin1000t` `[2COS^(2)THETA= (1+cos2theta)]` `y= 2sin1000t +2sin1000t.cost` `y= 2sin1000t+sin1001 t+sin999t` [as 2 sin A COS B = sin (A+B) + sin (A-B)] `:.` The given expression is the resultant of THREE independent simple harmonic motion. |
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| 12320. |
What is relative velocity? |
| Answer» Solution :When two objects are MOVING with DIFFERENT VELOCITIES, then the velocity of one object with RESPECT to another object is CALLED relative velocity of an object with respect to another. | |
| 12321. |
The moment of inertia of a seme circular ring of mass M and radius R about an axis lying in the plane which is passing through its centre and at an angle theta with the line joining its ends (as shown) is : |
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Answer» `("MR"^(2))/(4)" at "theta=0^(@)` |
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| 12322. |
What is the nature of the forces involved in an inelastic collision ? |
| Answer» SOLUTION :Some or all of the FORCES INVOLVED are NON conservative in NATURE. | |
| 12323. |
What is meant by Doppler effect ? Discuss the following case. (i) Source in motion and Observer at rest (a) Source moves towards observer (b) Source moves away from the observer (ii)Observer in motion and Source at rest. (a)bserver moves towards Source (b) Observer resides away from the Source (iii) Both are in motion (a) Source and Observer approach each other (b) Source and Observer resides from each other (c) Sourcechases Observer (d) Observer chases Source |
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Answer» Solution :When the source and the observer are in relative motion with respect to each other and to the medium in which sound in propagated, the frequency of the sound waves observed is different from the frequency of the source. This phenomeon is called Doppler Effect. Source in motion and the observer at rest. Source MOVES TOWARDS the observer.  Suppose a source S moves to the right (as shown in figure) with a velocity `v_(s)` and let the frequency of the sound waves produced by the source be `f_(s)`. it is assumed that the velocity of sound in a medium is v. The compression (sound wave front) produced by the source S at three successive instants of time are shown in the figure. When S is at position `x_(1)` the compression is at `C_(1)`. When s is at position `x_(2)` the compression is at `C_(2)` and similary for `C_(3) and C_(2)` . It is assumed that if `C_(1)` reaches the observer's position A then at the instant `C_(2)` reaches the point B and `C_(3)` reaches the point C as shown in the figure. Obvisouly it is seen that the distance between compresions, `C_(2) and C_(3)` is shorter than distance between `C_(1) and C_(2)` . It is meant by the wavelength decreases when the source S moves towards related to wavelength and therefore, frequency increases. Calculation : Let `LAMBDA` be the wavelength of the souce s as measured by the observer when S is at position `x_(1) and lambda` be wavelength of the source boserved by the observer when S moves to position `x_(2)`. Then the change in wavelength is `Delta lambda= lambda- lambda =v_(s)` t, where t is the time taken by the source to travel between `x_(1) and x_(2)`. Therefore, `lambda= lambda-vt ""....(1)` But `t=(lambda)/(v) ""...(2)` On substituting equations (2) in equation (1), we get. `lambda'= lambda (1-(v_(s))/(v))` Since frequency is inversely propprtional to wavelength, we have `f'=(v_(s))/(lambda') and f(v_(s))/(lambda)` Hence `f'=(f)/((1-(v_(s))/(v)))""...(3)` Since `(v_(s))/(v) lt lt 1`, by using the binomial expansion and retaining only first order in `(v_(s))/(v)` we get `f'=f(1+(v_(s))/(v)) ""....(4)` Source in motion and the observer at rest. Source moves away from the observer. Since the velocity of the source is opposite in DIRECTION when compared to case(a), hence by changing the sign of the velocity of the source in the above case i.,e by substituting `(v_(s) to -v_(s))` in equation (1), we get `f'=(f)/((1+(v_(s))/(v))) ""...(5)` Using binomial expansion again, we get `f'=f(1-(v_(s))/(v)) ""...(6)` Observer in motion and source at rest : Observer moves towards Source :  We can assume that the obsever O moves towards the source S with velocity, `v_(0)`. The source S is at rest and the velocity of sound wave (with respect to the medium) produced by the source is v. From the Figure, it is observed that both `v_(0) and v` arein opposite direction. Then, their relative velocity is `v_(r)=v+v_(0)`. The wavelenght of the sound is `v_(r)=v+v_(0)`. The wavelength of the sound wve is `lambda=(v)/(f)` which means the frequency observed by the observerO is `f'=(v_(r))/(lambda)`. Then `f'=(v_(r))/(lambda)=((v+v_(0))/(v))f` `=f(1+(v_(0))/(v))""...(7)` Observer in motion and source at rest : Observer receds away from the Source : If the observer O is moving away (receding away) from he source s, then velocity `v_(0)` andv moves in the same direction. Hence. their relative velocity is `v_(r)=v-v_(0)` Hence the frequency observerd by the observer O is `f'=(v_(r))/(lambda)=((v-v_(0))/(v))f` `=f(1-(v_(0))/(v)) ""...(8)` Both are in motion. : Both are in motion. : Source and observer approach each other  Let `v_(s) and v_(0)` be the respective velocities of source and observer approaching each other as shown in Figure. In order to calculate the apparent frequency observed by the obsever, let us have a dummy (behaving as observer or source) in between the source and observer. Since the dummy is at rest, dummy (observer) observes the apparent frequency due to approaching soure as given in equation `f'(f)/((1-(v_(s))/(v))` as. `f_(d)=(f)/((1-(v_(s))/(v))) ""....(1)` The true observer approaches the dummy from the other side at that instant of time. Since the source (true source) comes in a direction opposite to true observer, the dummy (source) is treated as stationary source for the true observer at that instant. Hence, apparent frequency when the true observer approaches the stationary source (dummy source), `f=f(1+(v_(0))/(v))` `f'=f_(d)(1+(v_(0))/(v))` `rArr f_(d)=(f)/((1+(v_(0))/(v))) ""....(2)` Since this is true any arbiitrary time, therefore, comparing equation (1) and equation (2), we get `(f)/((1-(v_(s))/(v)))=(f')/((1+(v_(0))/(v)))` `rArr(vf')/((v+v_(0)))=(vf)/((v-v_(s)))` Hence the apparent frequency as seen by the observer is `f'=((v+v_(0))/(v-v_(0)))f""...(3)` Both are in motion. : Source and observr recede from each other :  It is NOTICED that the velocity of the source and the observer each point in opposite directions with respect to the case in (a) and hence, we substitute `(v_(s) to -v_(s)) and (v_(0) to -v_(0))` in equation, (3), and therefore, the apparent frequency observed by the observer when the source and observer reced from each other is `f'=((v-v_(0))/(v+v_(s)))f` Both are in motion. : Source chases the obserer :  Only the observer's velocity is oppositely directed when compared to case (a) . Therefore, substituting `(v_(0) to -v_(0))` in equation (3), we get `f'=((v-v_(0))/(v-v_(s)))f` Observer chases the source :  Only the source velocity is oppositely directedwhen compared to case (a). therefore substituting `(v_(s) to -v_(s))` in equation (3), we get `f'=((v+v_(0))/(v+v_(s)))f` |
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| 12324. |
The ternimal velocity of small sized spherical ball of radius r falling vertically in a viscous liquid is |
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Answer» `alpha1/R^2` |
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| 12325. |
The sides of a rectangle are (5.5 +- 0.1)cm and (3.5 +- 0.2)cm. Its perimeter is |
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Answer» `(18.0 +- 0.1)CM` |
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| 12326. |
A particle executes SHM represented by the equation , y= 0.02sin(3.14t+ (pi)/(2)) meter. Find epoch |
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Answer» Solution :comparing equation `y= 0.02sin(3.14t+ (PI)/(2))`with the GENERAL from of the equation , `y= A sin(omegat+phi_(0))` Epoch `varphi= (pi)/(2)= (3.14)/(2)= 1.57 "RAD"` |
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| 12327. |
A system of binary stars of masses m_A and m_Bare moving is a circular orbits of radiusr_A .and r_Brespectively. If T_A and T_Bare the time periods of masses m_A and m_Brespectively then, |
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Answer» `T_A = T_B` `m_A = m_B r_B , T_A = T_B` |
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| 12328. |
The time taken for an ice block to slide down on inclined surface of inclination 60^(@) is 1-2 times the time taken by the same ice block to slide down a frictionless inclined plane of the same inclination. Calculate the coefficient of friction between ice and the inclined plane ? |
Answer» Solution :The force of friction `mu_(k)mg cos theta` opposes the component of theweight `mg sin theta` when theice BLOCK slides down.  Acceeration of ice `=a_(1)=g sin theta-mu_(k)gcostheta` `u=0` For the plane with friction, `s=ut+(1)/(2)at^(2)`, `s=0+(1)/(2)a_(1)t_(1)^(2)` For the frictionless plane, `a_(2)=g sin theta` `s=0+(1)/(2)a_(2)t_(2)^(2)` The distance TRAVELLED s is the same in the two cases `a_(1)t_(1)^(2)=a_(2)t_(2)^(2)` `t_(1)^(2)(gsintheta-mu_(k)gcostheta)=gsinthetat_(2)^(2)` `(t_(1)^(2))/(t_(2)^(2))=(gsintheta)/(gsintheta-mu_(k)gcostheta)` `(1.2)^(2)=(SIN60^(@))/(sin60^(@)-mu_(k)cos60)=(0.866)/(0.866-mu_(k)xx0.5)` `0.5mu_(k)=0.866-(0.866)/(1.44)=(0.265)/(0.5)=0.53` |
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| 12329. |
In dispersion without deviation |
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Answer» The emergent RAYS of all the colours are parallel to the incident RAY. |
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| 12330. |
A container of height 10m which is open at the top, has water to its full height. Two small openings are made on the walls of the container one exactly at the middle and the other at the bottom. The ratio of the velocities with which water comes out from the middle and the bottom region respectively is |
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Answer» 2  Velocity with which water COMING out from orifice 1, `v_(1) = SQRT(2g xx 5)` Velocity with which water coming out from orifice 2, `v_(2) = sqrt(2g xx 10)` `:. (v_(1))/(v_(2)) = sqrt((2g xx 5)/(2g xx 10)) = (1)/(sqrt2)` |
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| 12331. |
A copper wire and a steel wire of radii in the ratio 1:2, lengths in the ratio 2:1 are stretched by the same force. If the Young's modulus of copper = 1.1 xx 10^11Nm^(-2)find the ratio of their extensions (young's modulus of steel = 2 xx 10^11 N//m^2) . |
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Answer» Solution :we KNOW `e=(FL)/(pi R^(2)Y) RARR (e_(1))/(e_(2))=((L_(1))/(L_(2))) ((r_(2))/(r_(1)))^(2)((Y_(2))/(Y_(1)))((F)/(F))` Here `r_(1):r_(2)=1:2, L_(1):L_(2)=2:1, Y_(1)=1.1xx10^(11)Nm^(-2)`, `Y_(2)=2.0xx10^(11)Nm^(-2)` `(e_(1))/(e_(2))=(2)/(1)((2)/(1))^(2)((2.0xx10^(11))/(1.1xx10^(11)))=(16)/(1.1)=(160)/(11)` `e_(1):e_(2)=160:11` |
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| 12332. |
Explain method to determine size of molecule of oleic acid. |
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Answer» Solution :Oleic acid is a liquid like soap size of its molecular is of the order of `10^(-9) m`. In this method thickness of layer of molecular is determined. By using this DIMENSION of molecule is determined. `1 cm^(3)` oleic acid is mixed with `20 cm^(3)` alcohol. This is again mixed with 20 cm alcohol. Hence concentration of oleic acid per cm will be `((1)/(20xx20)) cm^(3)` This liquid is stored in a BROAD vessel and LYCOPODIUM POWDER is sprayed on it. A layer is formed on water surface. When oleic acid drop is made on surface it expand uniformly in circular shape and layer of one molecule is formed on it. To determine area (A) of layer of oleic acid diameter of layer is measured. Let n drop of oleic acid are formed on liquid (water) surface. Let volume of each drop `=V cm^(3)` `:.` Total volume of n drop = nV Volume of oleci acid in solution =nV `xx` concentration `=nvxx((1)/(20xx20)) cm^(3)` Let thickness of layer of oleic acid solution on water surface be t and area of layer be A then, `t=("volume of thin layer")/("area of layer")` `=(nV)/(20x20xxA) cm` If layer formed is one molecular layer then thickness t represent dimension of one molecule of oleic acid. Dimension (thickness) of oleic acid is of the order of `10^(-9) m` |
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| 12333. |
If the potential energy of the particle is alpha - (beta)/(2)x^(2) then force experienced by the particle is |
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Answer» `F=(BETA)/(2)X^(2)` |
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| 12334. |
The distance travelled by a body, falling freely from rest in t=1s, t=2s and t=3s are in the ratio of |
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Answer» `1:2:3` `:.Spropt^(2)` `:.S_(1):S_(2):S_(2)::1^(2):2^(2):3^(2)=1:4:9` |
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| 12335. |
S.T. v^2=(2gh)/(1+(K^2)/(R^2)) for the rolling object on an inclined plane of height (h) using dynamical Consideration. |
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Answer» Solution :Since energy is conserved `mgh=1/2mv^2+1/2Iomega^2` `mgh1/2mv^2+1/2mK^2(v^2)/(R^2)` `2gh=v^2+(K^2v^2)/(R^2)` or `2gh=v^2(1+(K^2)/(R^2))` where K=radius of GYRATION `v^2=(2gh)/((1+(K^2)/(R^2))` Note : for a CIRCULAR ring, k=R, For a circular disk K`=R/sqrt2`, for a solid cylinder `K=R/sqrt2`, for a hollow cylinder, `K=R`, For a solid sphere K=, and for a hollow sphere `K=(sqrt(2/3))R` |
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| 12336. |
When current passes through resistor due to its resistive property heat is produced. Error in measurement of resistance, current and time is 1%, 2% and 1% respectively, then what will be percentage error in measurement of heat energy ? |
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Answer» 4 `:.` Percentage error in measurement of heat energy, J=constant `=(DeltaH)/(H)xx100%=((2DeltaI)/(I)+(DELTAR)/(R)+(Deltat)/(t))xx100%` `=(2xx2)%+1%+1%` =6% |
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| 12337. |
In DeltaABC, mass of 100g, located on point A, mass of 200 g located on point C and mass of 150 g is located at point B, they are kept in xy-plane, find the coordinate of centre of mass of DeltaABC. |
| Answer» SOLUTION :`((7)/(36),(1)/(4sqrt(3)))` | |
| 12338. |
What is harmonic of oscilation (mode)? Give explanation of different harmonic (modes). |
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Answer» Solution :"The system of oscillations with natural frequency is called normal mode. " The possible minimum natural frequency is called fundamental mode of FIRST HARMONIC. Frequency for two ends of tensedstring of stationary wave, `v = (nv)/(2L)""...(1)` By substituing `v = (v)/(lamda)` `L = (nv lamda )/(2v) = (n lamda )/(2)""...(2)` and `lamda = (2L)/(n )""...(3)` where `n = 1,2,3,...` If `n =1,` then `v _(1) = (v )/(2L), L = (lamda _(1))/(2) and lamda _(1) = 2L` are obtained `v _(1)` is called fundamental or first harmonci. If `n =2,` then `v _(2) = (v)/(L) = 2v _(1),= lamda _(2) = lamda _(2) =L` are obtained `v _(2)` is called second harmonic. If `n =3,` then `v _(3) = (3v)/(2L) = 3v _(1), L = ( 3 lamda _(3))/(2) and lamda _(3) = (2)/(3)L` are obtained `v _(3)` is called third harmonic. Thus, by keeping `n = 4,5,6...` fourth, fifth sixth harmonic are obtained respectively which are given as below.  In figure, A = Antinode and N = node. Different harmonics can be represented by `v _(n) = nv _(1).` It is not necessary that string will oscillate by any one of the frequencies only. In general, string oscillates due to superposition of different modes. Among them some are strongly and some are weakly excited. Musical instruments like sitar and violin are designed on this principle. According to the principle of superposition, a stretched string tied at both END can vibrate simultaneously in more than one modes. Which mode is strongly excited DEPENDS on where the string is plucked or bowed. |
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| 12339. |
A uniform sphere of mass m and radius R starts rolling without slipping down an inclined plane. Find the time dependence of the angular momentum of the sphere relative to the point of contact at the initial moment. The angle of inclination of the plane is theta. How will the result be affected in the case of a perfectly smooth inclined plane 2/x (mgsintheta)Rt where 'x' is |
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| 12340. |
What value of temperature in farenheit scale becomes double than value in celcius scale ? |
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Answer» Solution :`T_(F)=(9)/(5)T_(C)+32,T_(F)2T_(C)` `:.2T_(C)=9/5T_(C)+32` `:.(T_(C))/(5)=32` `:.T_(C)=160^(@)C` `:.T_(F)=2T_(C)" GIVEN"`, `:.T_(F)=2xx160` `:.T_(F)=320" "^(@)F` |
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| 12341. |
Two particles move in a uniform gravitational field with an acceleration "g". At the initial moment the particles were located at same point and moved with velocities u_(1) = 0.8 ms^(-1) and u_(2) = 4.0 ms^(-1) horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular. (g = 10 ms^(-2)) |
Answer» 
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| 12342. |
Three rings, each of mass .M. and radius .R. are kept touching each other such that their centress from equilateral triangle. The M.I. of the system about a median of the triangle is |
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Answer» `3 MR^(2)` |
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| 12343. |
What is the angular velocity of a geostationary satellite if unit of time is an hour and that of angle is a degree ? |
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| 12344. |
If a wire is extended to a new length l, the work done is ......... |
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Answer» `(YA (l -l'))/(l)` `therefore F = (YA)/( l ) . Delta l =k. Delta l` `therefore` Force `PROP` extension If the extension is x, work done in extending by dx `dW = F. dx = KX dx` `therefore W = 1/2 kx ^(2) ` `If x is l .-l =x` `W = 1/2 k (l.-l)^(2)` MEANS work DONC `= 1/2 (YA)/(l) (l -1l.)^(2)` |
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| 12345. |
A uniform disc of mass M and radius R is pivoted about a horizontal axis passing through its edge. It is released from rest with its centre of mass at the same height as the pivot. |
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Answer» The ANGULAR velocity of disc when its centre of mass is directly below the pivot is `sqrt((4G)/(3R))` |
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| 12346. |
What are the characteristics of the ideal gas ? |
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Answer» SOLUTION :(i) The size of the MOLECULES is NEGLIGIBLY small . (ii) There is no force of attraction or REPULSION against its molecules. |
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| 12347. |
Can displacement and acceleration be in the same direction in SHM? |
| Answer» SOLUTION :No.in SHM,acceleration is always in the opposite of DISPLACEMENT. | |
| 12348. |
If the force of gravity acts on all bodies in proportion to their masses, why does a heavy body not fall faster than a light body? |
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Answer» SOLUTION :`F = Gmm/R^2` = MG but G = `GM/R^2` i.e.. g does not DEPEND on m. |
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| 12349. |
A body moves a distance of 20m along a straight line under the action of a force of 10N. If the work done is 100J, the angle between force and displacement vectors is |
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Answer» `0^(@)` |
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| 12350. |
A body proected at 45^(@) with a velocity of 20 m/s has a range of 10m. The decrease in range due to air resistance is (g = 10ms^(-1)) |
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Answer» 0 |
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