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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 39301. |
Water rises upto a height of 12 cm into a capillary tube when placed vertically. Ifit is now tilted through 45^(@) from the vertical, length of the water in the capillary tube is |
| Answer» <html><body><p>12 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a> <br/>`6sqrt(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)cm`<br/>`12sqrt(2)cm`<br/>6 cm</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 39302. |
The density of mercury is 13.6 gm/c.c at 0°C, and it is 13.15 gm/c.c at 130^(@)C. Calculate the coefficient of absolute expansion of mercury. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`26.32 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>) // ^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`</body></html> | |
| 39303. |
Find the expression of KE of the rigid body in rotational motion. |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C05_E02_124_S01.png" width="80%"/> <br/> Let us consider a rigid body rotating with angualar velocity `<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a>` about an axis as shown in the figure. Every particle of the bdoy will have the same <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> velocity `omega` and different <a href="https://interviewquestions.tuteehub.com/tag/tangential-1239000" style="font-weight:bold;" target="_blank" title="Click to know more about TANGENTIAL">TANGENTIAL</a> velocities v <a href="https://interviewquestions.tuteehub.com/tag/based-389387" style="font-weight:bold;" target="_blank" title="Click to know more about BASED">BASED</a> on its positions from the axis of rotation. <br/> Let us choose a particle of mass `m_(i)` situatedat distance `r_(i)` from the axis of rotation. It has a tangential velocity`v_(i)` given by the relation, `v_(i)=r_(i) omega`. The kinetic energy `KE_(i)` of the particle is <br/> `KE_(i)=1/2 m_(i)v_(i)^(2)` <br/> Since `v_(i)=r_(i)omega`<br/> `KE_(i)=1/2m_(i)(r_(i)omega)^(2)=1/2m_(i)(r_(i)^(2))omega^(2)` <br/> For the kinetic energy of the whole body, which is made up of large number of such particles, the equation is written with simmation as <br/> `KE=1/2(sum m_(i)r_(i)^(2))omega^(2)` <br/> where the term `sum m_(i)r_(i)^(2)` is the <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> of inertia I of the whole body. `I=sum m_(i)r_(i)^(2)` <br/> In rotational motion kinetic energy is <br/> `KE=1/2 I omega^(2)`</body></html> | |
| 39304. |
An object of mass 3 kg is at rest. Now a force of vecF=6t^(2)hati+4thatj is applied on the object, then the velocity of object at t=3 second is |
| Answer» <html><body><p>`18hati+3hatj`<br/>`18hati+6hatj`<br/>`3hati+18hatj`<br/>`18hati+4hatj`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`F=<a href="https://interviewquestions.tuteehub.com/tag/6t-332119" style="font-weight:bold;" target="_blank" title="Click to know more about 6T">6T</a>^(2)hati+4thatj` <br/> `F=ma"":.a=(<a href="https://interviewquestions.tuteehub.com/tag/barf-2460769" style="font-weight:bold;" target="_blank" title="Click to know more about BARF">BARF</a>)/(m)=(1)/(3)(6t^(2)hati+4thatj)=<a href="https://interviewquestions.tuteehub.com/tag/2t-301164" style="font-weight:bold;" target="_blank" title="Click to know more about 2T">2T</a>^(2)hati+(4)/(3)thatj` <br/> `a=(dV)/(dt)"":.v=int_(0)^(t)adt=int_(0)^(3)2t^(2)hatjdt+int_(0)^(3)(4)/(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)thatjdt=18hati+6hatj`</body></html> | |
| 39305. |
A body of mass 1 kg begins to move under the action of a dependent force vec(F)=(2t hat(i)+3t^(2)hat(j))N where hat(i) and hat(j) are unit vectors along x and y axis. What power will be developed by the force at the time t ? |
| Answer» <html><body><p>`(2t^(2)+<a href="https://interviewquestions.tuteehub.com/tag/3t-311358" style="font-weight:bold;" target="_blank" title="Click to know more about 3T">3T</a>^(3))<a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a>`<br/>`(2t^(2)+4t^(4))W`<br/>`(2t^(3)+3t^(4))W`<br/>`(2t^(3)+3t^(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))W` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 39306. |
The cross-sectional area of water pipe entering the basement is 4 x 10^(-4) m^(2). The pressure at this point is 3 x 10^(5) Nm^(-2) and the speed of water is 2 ms^(-1). This pipe tapers to a cross-sectional area of 2 x 10^(-4)m^(2) when it reaches the second floor 8 m above. Calculate the speed and pressure at the 2nd floor |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Since `A_(1)nu_(1)=A_(2)nu_(2)` <br/> `nu_(2)=(2xx4xx10^(-4))/(<a href="https://interviewquestions.tuteehub.com/tag/2xx10-1840192" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX10">2XX10</a>^(-4))` = 4m/s <br/> Using <a href="https://interviewquestions.tuteehub.com/tag/bernoulli-395852" style="font-weight:bold;" target="_blank" title="Click to know more about BERNOULLI">BERNOULLI</a>’s Theorem <br/> `P_(2)=P_(1)+(1)/(2)<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a>(upsilon_(1)^(2)-upsilon_(2)^(2))+rhog(h_(1)-h_(2))`<br/> `:. nu_(2)gtnu_(1)` <br/> `h_(2)gth_(1)` <br/> `=3xx10^(5)+(1)/(2)(1000)[(2)^(2)-(4)^(2)]-1000xx9.8xx8` <br/> `=2.16xx10^(5)N//m^(2)`</body></html> | |
| 39307. |
Which type of elastic modulus is there in liquid and gas ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> - <a href="https://interviewquestions.tuteehub.com/tag/elastic-967540" style="font-weight:bold;" target="_blank" title="Click to know more about ELASTIC">ELASTIC</a> <a href="https://interviewquestions.tuteehub.com/tag/modulus-1100097" style="font-weight:bold;" target="_blank" title="Click to know more about MODULUS">MODULUS</a> (<a href="https://interviewquestions.tuteehub.com/tag/bulk-403774" style="font-weight:bold;" target="_blank" title="Click to know more about BULK">BULK</a> modulus)</body></html> | |
| 39308. |
Explain how geocentric theory is required by helliocentric theory using the idea of retrograde motion of planets. |
| Answer» <html><body><p></p>Solution :(i) To explain this retrograde <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a>, Ptolemy introduced the concept of "epicycle" in his <a href="https://interviewquestions.tuteehub.com/tag/geocentric-2093615" style="font-weight:bold;" target="_blank" title="Click to know more about GEOCENTRIC">GEOCENTRIC</a> model. According to this theory, while the <a href="https://interviewquestions.tuteehub.com/tag/planet-1155613" style="font-weight:bold;" target="_blank" title="Click to know more about PLANET">PLANET</a> orbited the Earth, it also underwent <a href="https://interviewquestions.tuteehub.com/tag/another-876628" style="font-weight:bold;" target="_blank" title="Click to know more about ANOTHER">ANOTHER</a> circular motion termed as "epicycle". <br/> (ii) A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth. <br/> (iii) But Ptolemy's model became more and more complex as every planet was found to undergo retrograde motion. <br/> (iv) In the 15th century, the Polish astronmer Copernics proposed.<br/> (v) The helicentric model to explain this problem in a simpler manner. According to this model, the Sun is at the center of the solar system and all planets orbited the Sun. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/vi-723586" style="font-weight:bold;" target="_blank" title="Click to know more about VI">VI</a>) The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth.</body></html> | |
| 39309. |
In the above probllem, the normal force between the ball and the shell in position B is (m=mass of ball) |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)mg`<br/>`(7)/(9)mg`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/17-278001" style="font-weight:bold;" target="_blank" title="Click to know more about 17">17</a>)/(7)mg`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)/(7)mg`</p>Answer :C</body></html> | |
| 39310. |
Two particles A and B of equal mass M are moving with the same speed u as shown in the figure. They collide completely inelastically and move as a single particle C. The angle 9 that the path of C makes with the X-axis is given by : |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/tan-1238781" style="font-weight:bold;" target="_blank" title="Click to know more about TAN">TAN</a> theta =(<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(3) + sqrt(2))/(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>-sqrt(2))`<br/>`tan theta =(1+ sqrt(3))/(1+sqrt(2))`<br/>`tan theta =(sqrt(3) + sqrt(3))/(1-sqrt(2))`<br/>`tan theta =<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`</p>Answer :A</body></html> | |
| 39311. |
Compare the components for the following vector equations (a) Thatj-mghatj=mahatj (b) vecT+vecF=vecA+vecB © vecT-vecF=vecA-vecB (d) Thatj+mghatj=mahatj |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Components of the <a href="https://interviewquestions.tuteehub.com/tag/vectors-14090" style="font-weight:bold;" target="_blank" title="Click to know more about VECTORS">VECTORS</a> <br/> `T-mg=ma` <br/> (b) `barT_(x)+barF_(x)+barB_(x)` (or) `barT_(y)+barF_(y)=barA_(y)+barB_(y)` <br/> © `barT_(x)-barF_(x)+barB_(x)` (or) `barT_(y)-barF_(y)=barA_(y)+barB_(y)` <br/> (d) `T+mg=ma`</body></html> | |
| 39312. |
(n - 1) equal point masses each of mass m are placed at the vertices of a regular n-polygon. The vacant vertex has a position vector vecaw.r.t the centre of polygon. The position vector of centre of mass is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/veca-3257210" style="font-weight:bold;" target="_blank" title="Click to know more about VECA">VECA</a>/n`<br/>`veca/(1-n)`<br/>`(nveca)/(n-1)`<br/>`veca` </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 39313. |
Force of friction and tension in a string are |
| Answer» <html><body><p>Gravitational <a href="https://interviewquestions.tuteehub.com/tag/forces-16875" style="font-weight:bold;" target="_blank" title="Click to know more about FORCES">FORCES</a><br/>Electromagnetic forces<br/>Nuclear forces<br/>Weal forces</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> of friction and tension in a <a href="https://interviewquestions.tuteehub.com/tag/string-11290" style="font-weight:bold;" target="_blank" title="Click to know more about STRING">STRING</a> are electromagnetic forces.</body></html> | |
| 39314. |
In a system of units, the units of mass, length and time are 1 quintal, 1 km and 1h, respectively. In this system 1 N force will be equal to |
| Answer» <html><body><p>1 new <a href="https://interviewquestions.tuteehub.com/tag/unit-1438166" style="font-weight:bold;" target="_blank" title="Click to know more about UNIT">UNIT</a> <br/>129.6 new unit <br/>427.6 new unit <br/>60 new unit </p>Solution :`["Force"]=["MLT"^(-<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)]` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a>" "1"<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>"=((1)/(100))((1)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>))(3600)^(2)=129.6" units"`.</body></html> | |
| 39315. |
The displacement of a particle moving along x-axis with respect to time t is x = at + bt^(2) - ct^(3)The dimensions of care |
| Answer» <html><body><p>`[T^(-3)]`<br/>`[LT^(-2)]`<br/>`[LT^(-3)]`<br/>`[LT^(3)]`</p>Solution :`x= at+ bt^(2)-<a href="https://interviewquestions.tuteehub.com/tag/ct-411317" style="font-weight:bold;" target="_blank" title="Click to know more about CT">CT</a>^(3)` <br/> Each term on <a href="https://interviewquestions.tuteehub.com/tag/rhs-614248" style="font-weight:bold;" target="_blank" title="Click to know more about RHS">RHS</a> <a href="https://interviewquestions.tuteehub.com/tag/must-2185568" style="font-weight:bold;" target="_blank" title="Click to know more about MUST">MUST</a> have dimensions of x.<br/> `:. [ct^(3)]= [x]` or `[c]= ([x])/([t^(3)])= ([<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>])/([T^(3)])= [LT^(-3)]`</body></html> | |
| 39316. |
When a train of plane wave traverses a medium, individual particles execute periodic motion given by the equation y 4sin(pi)/(2)(2t x/8) Where the length are expressed in centimetres and time in seconds. Calculate the amplitute, wavelength, (a) the phase different for two positions of the same particle which are occupied at time interval 0.4 s apart and (b) the phase difference at any given instant of two particle 12 cm apart. |
| Answer» <html><body><p></p>Solution :The equation of a wave <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a> is given by<br/> `y=A <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a>(2pi)/(lambda)(vt+x)`(i) <br/> Here, <br/> `y 4 sin(pi)/(2)(2t x/8)`<br/> This equation canbe written as <br/> `y 4sin(2pi)/(32)(16t x)` (ii) <br/> comparing `Eq`. (i)with `Eq`. (ii), we get <br/> Amplitude `A=4 <a href="https://interviewquestions.tuteehub.com/tag/cm-919986" style="font-weight:bold;" target="_blank" title="Click to know more about CM">CM</a>,`wavelength `(lambda)=32 cm`, wave velocity `v=16 cm//s` <br/> here frequency is given as <br/> `f=(v)/(lambda)=(16)/(32)=(1)/(2)=0.5 <a href="https://interviewquestions.tuteehub.com/tag/hz-493442" style="font-weight:bold;" target="_blank" title="Click to know more about HZ">HZ</a>` <br/> (a). phase of a particle at <a href="https://interviewquestions.tuteehub.com/tag/instant-1046099" style="font-weight:bold;" target="_blank" title="Click to know more about INSTANT">INSTANT</a> `t_(1)` is given by<br/> `phi_(1) (pi)/(2)(2t_(1) x/8)` <br/> The phase at instant `t_(2)` is given by `phi_(2) (pi)/(2)(2t_(1) x/8)` <br/> The phase difference is given as <br/> `phi_(1)-phi_(2)=(pi)/(2)[(2t_(1)+x/8)-(2t_(2)+x/8)]` <br/> `pi(t_(1)-t_(2)) pi(0.4)` `(As t_(1)-t_(2)=0.4)` <br/> `=180xx0.4=72^@``(pi rad=180^@)` <br/> (b). phase different at an instant between two particle with path different `(Delta)` is<br/> `phi=(2pi)/(lambda)xxDelta``(2pi)/(32)xx12``(As Delta=12 cm` <br/> `(3pi)(4)`</body></html> | |
| 39317. |
A hot water cools from 92^(@)C to 84^(@)C in 3 minutes when the room temperature is 27^(@)C. How long will it take for it to cool from 65^(@)C to 60^(@)C? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/hot-1029680" style="font-weight:bold;" target="_blank" title="Click to know more about HOT">HOT</a> water cools `<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>^(@)C` in 3 minutes. The average temperature of `92^(@)C` and `84^(@)C` is `<a href="https://interviewquestions.tuteehub.com/tag/88-339824" style="font-weight:bold;" target="_blank" title="Click to know more about 88">88</a>^(@)C`. This average temperature is `61^(@)C` above room temperature. By using equation, <br/> `(dT)/(T-T_(s))=-(a)/(<a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>)dt or (dT)/(dt)=-(a)/(ms)(T-T_(s))`<br/> `(8^(@)C)/("3 min")=-(a)/(ms)(61^(@)C)"...(1)"` <br/> Similarly the average temperature of `<a href="https://interviewquestions.tuteehub.com/tag/65-331005" style="font-weight:bold;" target="_blank" title="Click to know more about 65">65</a>^(@)C and 60^(@)C`. The average temperature is `35.5^(@)C` above the room temperature. Then we can write <br/> `(5^(@)C)/(dt)=-(a)/(ms)(35.5^(@)C)"....(2)"` <br/> By dividing both the equation, we get <br/> `((8^(@)C)/("3 min"))/((5^(@)C)/(dt))=-((a)/(ms)(61^(@)C))/(-(a)/(ms)(35.5^(@)C))` <br/> `(8xxdt)/(3xx5)=(61)/(35.5)` <br/> `dt=(61xx15)/(35.5xx8)=(915)/(284)="3.22 min"`</body></html> | |
| 39318. |
If the velocity of light c, Planks constant, h and the gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities. |
| Answer» <html><body><p></p>Solution :Here, `c=[LT^-1],h=[ML^2T^-1],`<br/>`G=[M^-1L^3T^-2]`<br/>`becauseE=hv,h=E/v,G=((Fd^2)/(m_1m_2))`<br/>Let `m=c^xh^yG^zto(1)`<br/>`implies[M^1L^0T^0]=(LT^-1)^x(ML^2T^-1)^y(M^-1L^3T^-2)^<a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a>`<br/> `implies[M^1L^0T^0]=M^(y-z)L^(x+2y+2z)T^(-x-y-2z)`<br/>Applying the principle of homegenity of <a href="https://interviewquestions.tuteehub.com/tag/dimensions-439808" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONS">DIMENSIONS</a>, we get <br/> `y-z=1to(2),x+2y+3z=0to(3),`<br/> `-x-y-2z=0to(4)`<br/>Adding <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>.(2), eq.(3) and eq.(4). <br/> `2y=1impliesy=1/2` <br/> `therefore` From eq.(2) `z=y-1=1/2-1=(-1)/2`<br/>From eq.(4) `x=-y-2z=(-1)/2+1=1/2` <br/> Substituting the values of x,y & z in eq.(1) , we get <br/> `m=c^(1//2)h^(1//2)G^(-1//2)impliesm=sqrt((<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>)/G)`<br/>Proceeding as above we can show that <br/> `L=sqrt((hG)/c^3)andT=sqrt((hG)/c^5)`</body></html> | |
| 39319. |
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t = 2. The gap is found to remain constant. The velocities with which the balls were thrown are (Take g = 10 m s^(-1)) |
| Answer» <html><body><p>`20 m s^(-1) , 10 m s^(-1)`<br/>`10 m s^(-1) , 5 m s^(-1)`<br/>`16 m s^(-1) , 8 m s^(-1)`<br/>`30 m s^(-1), 15 m s^(-1)`</p>Solution :For first <a href="https://interviewquestions.tuteehub.com/tag/stone-1228007" style="font-weight:bold;" target="_blank" title="Click to know more about STONE">STONE</a>, <br/> taking the vertical upwards motion of the first stone up to highest point <br/> Here, u = `u_1`, v = 0 (At highest point velocity is zero) <br/> a = -g, S = `h_1` <br/> As `v^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) - u^(2) = 2aS``<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> (0)^(2) - u_(1)^(2) = 2(-g)h_1`or`h_1 = u_(1)^(2)/(2g)`...(i)<br/> For second stone, <br/> Taking the vertical upwards motion of the second stone up to highest point <br/> here, u = `U_2, v = 0, a = -g, S = h_2` <br/> As `v^(2) - u^(2)` = 2as <br/> `therefore(0)^(2) - (u_2)^(2) = 2(-g)h_2`or`h_2 = u_(1)^(2)/(2g)`.............(ii) <br/> As per question <br/> `H_1 - h_2 = 15 m , u_2 = u_1/2` <br/> <a href="https://interviewquestions.tuteehub.com/tag/subtract-1231765" style="font-weight:bold;" target="_blank" title="Click to know more about SUBTRACT">SUBTRACT</a> (ii) from (i), we get, `h_1 - h_2 = u_(1)^(2) /(2g)-(u_(2)^(2)/(2g))`<br/> On substituting the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> information, we get <br/> `15 = u_(1)^(2)/(2g)-u_(1)^(2)/(2g)=(3u_(1)^(2))/(8g)` or `u_(1)^(2)=(15 xx 8g)/(3)=(15 xx 8 xx 10)/(3)= 400` <br/> or`u_1 = 20 m s^(-1)` and`u_2 = U_1/2 = 10 m s^(-1)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XI_C03_E01_065_S01.png" width="80%"/></body></html> | |
| 39320. |
Pitch of sound depends on ……….. . |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/wavelength-1450414" style="font-weight:bold;" target="_blank" title="Click to know more about WAVELENGTH">WAVELENGTH</a> <br/>amplitude<br/>speed </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :frequency</body></html> | |
| 39321. |
Temperature of a body theta is slightly more than the temperature of the surroundings theta_(0). Its rate of cooling ( R ) versus temperature of body (theta) is plotted, its shape would be : |
| Answer» <html><body><p><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_PHY_XI_V01_D_C04_E01_083_O01.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_PHY_XI_V01_D_C04_E01_083_O02.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_PHY_XI_V01_D_C04_E01_083_O03.png" width="30%"/> <br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_PHY_XI_V01_D_C04_E01_083_O04.png" width="30%"/> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 39322. |
Explain how Newton derived his law of gravitation from Kepler's third law. |
| Answer» <html><body><p></p>Solution :Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the centre is <br/>`a=-(v^2)/(r)""...(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C06_E01_032_S01.png" width="80%"/> <br/>Here v is the velocity and r, the distance of the planet from the centre of the orbit.<br/>The velocity in <a href="https://interviewquestions.tuteehub.com/tag/terms-1242559" style="font-weight:bold;" target="_blank" title="Click to know more about TERMS">TERMS</a> of known quantities r and T, is <br/>`v=(2pir)/(T)""...(2)` <br/>Here T is the time period of <a href="https://interviewquestions.tuteehub.com/tag/revolution-623158" style="font-weight:bold;" target="_blank" title="Click to know more about REVOLUTION">REVOLUTION</a> of the planet. Substituting the value of 'v' in equation (1) we get ,<br/>`a=-(((2pir)/(T))^(2))/(r)` <br/>`=-(<a href="https://interviewquestions.tuteehub.com/tag/4pi-1882352" style="font-weight:bold;" target="_blank" title="Click to know more about 4PI">4PI</a>^(2)r)/(T^2)""...(3)` <br/>Substituting the value of 'a' from (3) in Newton's second law, F = ma where 'm' is the mass of the planet.<br/>`F=-(4pi^(2)mr)/(T^2)""...(4)` <br/>From Kepler's third law,<br/>`(r^3)/(T^2)=k("constant")""...(5)` <br/>`(r)/(T^2)=(k)/(r^2)""...(6)` <br/>By equation (6) in the force expression, we can arrive at the law of gravitation.<br/>`F=-(4pi^(2)mk)/(r^2)""...(7)` <br/>Here negative sign implies that the force is <a href="https://interviewquestions.tuteehub.com/tag/attractive-887662" style="font-weight:bold;" target="_blank" title="Click to know more about ATTRACTIVE">ATTRACTIVE</a> and it acts towards the centre.He equated the constant `4pi^(2)k` to GM which turned out to be the law of gravitation.<br/>`F=-(GMm)/(r^2)`</body></html> | |
| 39323. |
Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_BEN_PHY_XI_P2_C14_E01_003_O01.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_BEN_PHY_XI_P2_C14_E01_003_O02.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_BEN_PHY_XI_P2_C14_E01_003_O03.png" width="30%"/><br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_BEN_PHY_XI_P2_C14_E01_003_O04.png" width="30%"/><br/></p>Solution :(b) and (d) are periodic, each with a <a href="https://interviewquestions.tuteehub.com/tag/period-1151023" style="font-weight:bold;" target="_blank" title="Click to know more about PERIOD">PERIOD</a> of <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> s, (a) and (c) are not periodic. [Note in (c), repetition of <a href="https://interviewquestions.tuteehub.com/tag/merely-7377077" style="font-weight:bold;" target="_blank" title="Click to know more about MERELY">MERELY</a> <a href="https://interviewquestions.tuteehub.com/tag/one-585732" style="font-weight:bold;" target="_blank" title="Click to know more about ONE">ONE</a> position is not <a href="https://interviewquestions.tuteehub.com/tag/enough-446095" style="font-weight:bold;" target="_blank" title="Click to know more about ENOUGH">ENOUGH</a> for motion to be periodic, the entire motion during one period must be repeated successively].</body></html> | |
| 39324. |
A constant power is supplied to a rotating disc. The relationship between the angular velocity ( omega) of the disc and number of rotations (n) made by the disc is governed by |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/omega-585625" style="font-weight:bold;" target="_blank" title="Click to know more about OMEGA">OMEGA</a> <a href="https://interviewquestions.tuteehub.com/tag/prop-607409" style="font-weight:bold;" target="_blank" title="Click to know more about PROP">PROP</a> <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>^(1//3)`<br/>`omega prop n^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>//3)`<br/>`omega prop n^(3//2)`<br/>`omega prop n^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 39325. |
Match list I with list II LIST - I(a) Centrifugal force(b) Centripetal force(c ) Tangential force (d) Angular velocityLIST - II(e ) Along the axis of rotation (f) Towards the centre of rotation(g) Away from the centre of rotation(h) Changes the angular velocity |
| Answer» <html><body><p>a-h, <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>-g, c-f, d-e <br/>a-g, b-f, c-h, d-e<br/>a-f, b-g, c-h, d-e <br/>a-c, b-h, c-e, d-f</p>Answer :B</body></html> | |
| 39326. |
A 2kg stone is swung in a vertical circle by attaching it at the end of a string of length 2m. If the stringcan with stand a tension 140.6N, the maximum speed with which the stone can be rotated is |
| Answer» <html><body><p>`22ms^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)`<br/>`44ms^(-1)`<br/>`33ms^(-1)`<br/>`11ms^(-1)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 39327. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 6 cm away from B going towards A |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_013_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> is negative, <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> and force are <a href="https://interviewquestions.tuteehub.com/tag/positive-1159908" style="font-weight:bold;" target="_blank" title="Click to know more about POSITIVE">POSITIVE</a></body></html> | |
| 39328. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 3 cm away from A going towards B |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_012_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a>, <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> and <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> all are <a href="https://interviewquestions.tuteehub.com/tag/positive-1159908" style="font-weight:bold;" target="_blank" title="Click to know more about POSITIVE">POSITIVE</a></body></html> | |
| 39329. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 2 cm away from B going towards A, |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_011_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a>, <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> and <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> all are <a href="https://interviewquestions.tuteehub.com/tag/negative-570381" style="font-weight:bold;" target="_blank" title="Click to know more about NEGATIVE">NEGATIVE</a></body></html> | |
| 39330. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the mid point of AB going towards A, |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_010_S01.png" width="80%"/> <br/> At the mid <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> of AB (at mean position .O.) velocity is <a href="https://interviewquestions.tuteehub.com/tag/maximum-556915" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMUM">MAXIMUM</a> and negative as the particle <a href="https://interviewquestions.tuteehub.com/tag/moves-1104598" style="font-weight:bold;" target="_blank" title="Click to know more about MOVES">MOVES</a> towards A. acceleration and force are <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a></body></html> | |
| 39331. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end B, |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_009_S01.png" width="80%"/> <br/> At the end .B. <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>, <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> and <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> are negative</body></html> | |
| 39332. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end A, |
| Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_B_C08_SLV_008_S01.png" width="80%"/> <br/> At the end .A. particle is at extreme position hence velocity is zero. As it is acted up on by <a href="https://interviewquestions.tuteehub.com/tag/restoring-1187260" style="font-weight:bold;" target="_blank" title="Click to know more about RESTORING">RESTORING</a> <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> <a href="https://interviewquestions.tuteehub.com/tag/towards-7269729" style="font-weight:bold;" target="_blank" title="Click to know more about TOWARDS">TOWARDS</a> mean position .O. its <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> and force <a href="https://interviewquestions.tuteehub.com/tag/acting-269" style="font-weight:bold;" target="_blank" title="Click to know more about ACTING">ACTING</a> on it are positive</body></html> | |
| 39333. |
State the law of floatation. |
| Answer» <html><body><p></p>Solution :The law of floatation <a href="https://interviewquestions.tuteehub.com/tag/states-1225920" style="font-weight:bold;" target="_blank" title="Click to know more about STATES">STATES</a> that a body will <a href="https://interviewquestions.tuteehub.com/tag/float-992844" style="font-weight:bold;" target="_blank" title="Click to know more about FLOAT">FLOAT</a> in a liquid if the weight of the liquid displaced by the <a href="https://interviewquestions.tuteehub.com/tag/immersed-1037461" style="font-weight:bold;" target="_blank" title="Click to know more about IMMERSED">IMMERSED</a> <a href="https://interviewquestions.tuteehub.com/tag/part-596478" style="font-weight:bold;" target="_blank" title="Click to know more about PART">PART</a> of the body <a href="https://interviewquestions.tuteehub.com/tag/equals-974070" style="font-weight:bold;" target="_blank" title="Click to know more about EQUALS">EQUALS</a> the weight of the body.</body></html> | |
| 39334. |
. (A): Comets move around the sun in elliptical orbits. The gravitational force on the comet due to sun is not normal to the comet.s velocity but the work done by the gravitational force over every complete orbit of the comet is zero (R): Gravitational force is a non conservative force. |
| Answer» <html><body><p>Both .A. and .<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>. are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and .R. is the correct explanation of .A.<br/>Both .A. and .R. are true and .R. is not the correct explanation of .A. <br/>A. is true and .R. is false<br/> .A. is false and .R. is true</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 39335. |
Which of the following are positive and which are negative work done ? Work done by friction on a body sliding down an inclined plane. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/negative-570381" style="font-weight:bold;" target="_blank" title="Click to know more about NEGATIVE">NEGATIVE</a></body></html> | |
| 39336. |
In the Searle's method to determine the Young's modulus of a wire, a steel wire of length 156 cm and diameter 0.054 cm is taken as experimental wire. The average increase in length for 1 1/2 kg wt is found to be 0.050 cm. Then the Young's modulus of the wire. |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>.002xx10^(<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>)N//m^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <br/>`1.002xx10^(11)N//m^(2)`<br/>`2.002xx10^(11)N//m^(2)` <br/>`2.5xx10^(11)N//m^(2)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :3</body></html> | |
| 39337. |
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined.Consider a drop of mass 1.00 g falling from a hwight 1.00 km .It hits the groundwith a speed of 50.0 ms^(-1) (a) What is the work done gravitational force ? (b) What is the work done by the unknown resistive force ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(a) Wg = 20 <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> , <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `Wr = - 17.55`</body></html> | |
| 39338. |
Define vibratory motion. Give example |
| Answer» <html><body><p></p>Solution :If an object or particle executes a to-and-fro motion about a fixed point then it is said to bein vibratiory motion . This is sometimes also called <a href="https://interviewquestions.tuteehub.com/tag/oscillation-1139998" style="font-weight:bold;" target="_blank" title="Click to know more about OSCILLATION">OSCILLATION</a> . <a href="https://interviewquestions.tuteehub.com/tag/eg-445433" style="font-weight:bold;" target="_blank" title="Click to know more about EG">EG</a>: Vibration of a string on a <a href="https://interviewquestions.tuteehub.com/tag/guitar-1014020" style="font-weight:bold;" target="_blank" title="Click to know more about GUITAR">GUITAR</a> .</body></html> | |
| 39339. |
(A): In the pressure equation P=(1)/(3)av_(rms)^(2),the term 'a' represents density of gas (R ):rms velocity of gas v_(rms)=sqrt((3RT)/(M)) |
| Answer» <html><body><p>Both (A) and ( R) are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and (R ) is the <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of (A)<br/>Both (A) and (R ) are true and (R) is not the correct explanation of (A) <br/>(A) is true but (R ) is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a> <br/>Both (A) and (R ) are false </p>Answer :B</body></html> | |
| 39340. |
A scooter can produce a maximum acceleration of 5ms^(-2). Its brakes can produce a maximum retardation of 10ms^(-2). The minimum time in which it can cover a distance of 1.5 km is? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP1_C03_SLV_008_S01.png" width="80%"/> <br/> If v is the maximum velocity attained, <br/> then during acceleration is between A, B <br/> `v^(2)-o^(2)=2xx5xxS_(1)impliesS_(1)=(v^(2))/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)`, <br/> Also, during <a href="https://interviewquestions.tuteehub.com/tag/retardation-1187717" style="font-weight:bold;" target="_blank" title="Click to know more about RETARDATION">RETARDATION</a> <br/> `o^(2)-v^(2)=2xx10xxS_(2)impliesS_(2)=(v^(2))/(20)` <br/> `S=S_(1)+S_(2)implies1500=(v^(2))/(10)+(v^(2))/(20)=(3v^(2))/(20)or` <br/> `v^(2)=(1500xx20)/(3)=10000 or v=100ms^(-1)` <br/> `v=alphat_(1)impliest_(1)=(100)/(5)=<a href="https://interviewquestions.tuteehub.com/tag/20sec-293103" style="font-weight:bold;" target="_blank" title="Click to know more about 20SEC">20SEC</a>` <br/> `v=betat_(2)impliest_(2)=(100)/(10)=10sec` <br/> Total time `=20+10=30sec`.</body></html> | |
| 39341. |
A ball of mass 50 g falls from rest vertically downwards through a distance of 40 m and hits the ground. Find the kinetic energy and final velocity of the ball before it hits the ground (g = 9.8 ms^(-2)). |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> of ball, m = 50 gm `= 50 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-3)` kg , Height of fall, h = 40 m , g = 9.8 `ms^(-2)` <br/> From law of conservation of energy, loss in P.E = gain in K.E. <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> mgh = (1)/(2) <a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a>^(2) and v = sqrt(2 gh)` <br/> K.E. = mgh `rArr` K.E. = `50 xx 10^(-3) xx 9.8 xx 40 = 19.6` J <br/> Final <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a>, `v = sqrt(2 xx 9.8 xx 40) = 28 ms^(-1)`</body></html> | |
| 39342. |
What is the Poisson's ratio of the substance whose volume remains unchanged under elastic strains? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`1/2`</body></html> | |
| 39343. |
The graphical variation of the readings of an arbitrary scale X and Reaumer scale corresponding to the same temperatures. The temperature in Kelvin scale when the readings of the two scales coincide will be |
| Answer» <html><body><p>323<br/>313<br/>303<br/>333</p>Answer :A</body></html> | |
| 39344. |
An automobile that is towing a trailer is accelerating on a level road. The force that the automobile exerts on the trailer is |
| Answer» <html><body><p>equal to the force the <a href="https://interviewquestions.tuteehub.com/tag/trailer-711816" style="font-weight:bold;" target="_blank" title="Click to know more about TRAILER">TRAILER</a> <a href="https://interviewquestions.tuteehub.com/tag/exerts-979377" style="font-weight:bold;" target="_blank" title="Click to know more about EXERTS">EXERTS</a> on the automobile <br/>greater than the force the trailer exerts on the automobile <br/>equal to the force the trailer exerts on the <a href="https://interviewquestions.tuteehub.com/tag/road-1190248" style="font-weight:bold;" target="_blank" title="Click to know more about ROAD">ROAD</a> <br/>equal to the force the road exerts on the trailer </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 39345. |
A body is projected with a velocity 50ms^(-1). Distance travelled in 6^(th) second is [g=10ms^(-2)] |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> m <br/><a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> m<br/>15 m<br/>20 m</p>Answer :A</body></html> | |
| 39346. |
If reflecting and transmitting powers of a body are 0.2 and 0.3 units, then its Absorptive power will be |
| Answer» <html><body><p>0.1<br/>0.5<br/>0.25<br/>1</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 39347. |
A capilary tubes of radius R is dipped vertically in a liquid with surface tension T. If the rise of the liquid in the tube is h then the weight of the liquid column in the tube is |
| Answer» <html><body><p>`2piRT`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/pirt-3772500" style="font-weight:bold;" target="_blank" title="Click to know more about PIRT">PIRT</a>`<br/>`2piR^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)T`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/pit-1155063" style="font-weight:bold;" target="_blank" title="Click to know more about PIT">PIT</a>`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 39348. |
The mass of moon is 1% of mass of earth. The ratio of gravitational pull of earth on moon and that of moon on earth will be |
| Answer» <html><body><p>`1:1`<br/>`1:10`<br/>`1:100`<br/>`2:1`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 39349. |
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 xx 10 ^(8)Pa. A steel ball of initial volume 0.32 m^(3) is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom ? |
| Answer» <html><body><p></p>Solution :`P =1.1 xx 10 ^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) pa` <br m=""/> `B =1.6 xx 10 ^(<a href="https://interviewquestions.tuteehub.com/tag/11-267621" style="font-weight:bold;" target="_blank" title="Click to know more about 11">11</a>) <a href="https://interviewquestions.tuteehub.com/tag/nm-579234" style="font-weight:bold;" target="_blank" title="Click to know more about NM">NM</a> ^(-2)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/bulk-403774" style="font-weight:bold;" target="_blank" title="Click to know more about BULK">BULK</a> modulus `B = (P)/((Delta V)/(V))` (<a href="https://interviewquestions.tuteehub.com/tag/magnitude-1083080" style="font-weight:bold;" target="_blank" title="Click to know more about MAGNITUDE">MAGNITUDE</a>) <br/> `B = (PV)/(Delta V) ` <br/> `therefore Delta V =(PV)/(B) = (1.1 xx 10 ^(8) xx 0.3)/(1.6 xx 10 ^(11))=22 x 10 ^(-5)` <br/> `~~ 2.2 xx 10 ^(-4) m ^(3)` <br/> Notye : Answer of Text-book is not matching, if for steel B = 140 GPa, then the answer of Text-book will be `Delta V =2.51 xx 10 ^(-4) m ^(3).`</body></html> | |
| 39350. |
The density of water is more than the density of air . Even then clouds containing water droplets don’t fall and continue to float. Explain. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :At terminal velocity `v_T (alpha r^2)`, dur to <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> size of droplets in the <a href="https://interviewquestions.tuteehub.com/tag/cloud-11874" style="font-weight:bold;" target="_blank" title="Click to know more about CLOUD">CLOUD</a>, `v_T` is very small, so the cloud falls very <a href="https://interviewquestions.tuteehub.com/tag/slowly-2273700" style="font-weight:bold;" target="_blank" title="Click to know more about SLOWLY">SLOWLY</a> towards earth and appears to float in the <a href="https://interviewquestions.tuteehub.com/tag/sky-1211556" style="font-weight:bold;" target="_blank" title="Click to know more about SKY">SKY</a>.</body></html> | |