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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.
| 37951. |
We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ? |
| Answer» <html><body><p></p>Solution :Because, for an arbitrarily <a href="https://interviewquestions.tuteehub.com/tag/small-1212368" style="font-weight:bold;" target="_blank" title="Click to know more about SMALL">SMALL</a> <a href="https://interviewquestions.tuteehub.com/tag/interval-1049953" style="font-weight:bold;" target="_blank" title="Click to know more about INTERVAL">INTERVAL</a> of time, the magnitude of displacement is equal to the <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a> of the <a href="https://interviewquestions.tuteehub.com/tag/path-11833" style="font-weight:bold;" target="_blank" title="Click to know more about PATH">PATH</a>.</body></html> | |
| 37952. |
Find the torque of a force 7hati + 3hatj – 5hatkabout the origin. The force acts on a particle whose position vector is hati – hatj + hatk. |
| Answer» <html><body><p></p>Solution :Here `r=<a href="https://interviewquestions.tuteehub.com/tag/hati-2693493" style="font-weight:bold;" target="_blank" title="Click to know more about HATI">HATI</a>-hatj+hatk` <br/> and `F=7hati+3hatj-5hatk` <br/> We <a href="https://interviewquestions.tuteehub.com/tag/shall-641229" style="font-weight:bold;" target="_blank" title="Click to know more about SHALL">SHALL</a> use the determinant rule to <a href="https://interviewquestions.tuteehub.com/tag/find-11616" style="font-weight:bold;" target="_blank" title="Click to know more about FIND">FIND</a> the torque `tau=rxxF` <br/> `tau=|(hati,hatj,hatk),(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>,-1,1),(7,<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>,-5)|=(5-3)hati-(-5-7)hatj+(3-(-7))hatk` <br/> or `tau=2hati+12hatj+10hatk`</body></html> | |
| 37953. |
Two identical rods of a metal are welded in series then 20 cal of heat flows through them in 4 minute. If the rods are welded in parallel then the same amount of heat will flow in |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> minute<br/>2 minute<br/>4 minute<br/>16 minute</p>Answer :A</body></html> | |
| 37954. |
Calculate rate of flow of glycerin of density 1.25xx10^(3) kg/m^(3) through the concial section of a horizontal pipe if the radii of its ends are 0.1 m and 0.04m and pressure drop across its length is 10 N//m^(2). |
| Answer» <html><body><p></p>Solution :According to <a href="https://interviewquestions.tuteehub.com/tag/continuity-932023" style="font-weight:bold;" target="_blank" title="Click to know more about CONTINUITY">CONTINUITY</a> equation `(v_(2))/(v_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)` <br/>and, according to Bernouli.s equation for a horizontal tube `P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)` <br/>`V_(2)^(2)-V_(1)^(2)=2(P_(1)-P_(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>)=2xx(10N//m^(2))/((1.25xx10^(3)kg//m^(3)))=16xx10^(-3)m^(2)//s^(2)` but `v_(2)=(25)/(4)v_(1)=6.25v_(1)` <br/>`therefore [(6.25)^(2)-t^(2)]V_(1)^(2)=16xx10^(-3)m^(2)//s^(2)` or `V_(1)~~0.0205m//s` <br/>So,the rate of <a href="https://interviewquestions.tuteehub.com/tag/volume-728707" style="font-weight:bold;" target="_blank" title="Click to know more about VOLUME">VOLUME</a> flow=`A_(1)V_(1)=<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s` <br/>And the rate of mass flow is `(dm)/(dt)=rhoAV=(1.25xx10^(3)kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785 kg//s`</body></html> | |
| 37955. |
A thin uniform rod of mass 'm' and length 'l' is kept on a smooth horizontal surface such that it can move freely. Where should a particle of mass 'm' moving horizontally strike on the rod from the centre (x) such that the point 'A' at a distance l/3 from the end of the rod is instantaneously at rest just after the elastic collision ? |
| Answer» <html><body><p>l/2<br/>l/3<br/>l/3<br/>l/3</p>Answer :A</body></html> | |
| 37956. |
A satellite orbits the earth at a height of R/6, its orbital speed ? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((2GM)/(<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>))`<br/>`sqrt((<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/(R))`<br/>`sqrt((GM)/(7R))`<br/>`sqrt((6GM)/(7R))`</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> | |
| 37957. |
What is the condition to be satisfied by a mathematical relation between time and displacement to describe a periodic motion ? |
| 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 perodic <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a> <a href="https://interviewquestions.tuteehub.com/tag/repeats-1185149" style="font-weight:bold;" target="_blank" title="Click to know more about REPEATS">REPEATS</a> after a definite <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> <a href="https://interviewquestions.tuteehub.com/tag/interval-1049953" style="font-weight:bold;" target="_blank" title="Click to know more about INTERVAL">INTERVAL</a> T. So, `y(t) = y(t + T)=y(t+2T)` etc.</body></html> | |
| 37958. |
A spring with 10 coils has spring constant k. It is exactly cut into two halves, then each of these new springs will have a spring constant |
| Answer» <html><body><p>`k//2`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/3k-310983" style="font-weight:bold;" target="_blank" title="Click to know more about 3K">3K</a>//2`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2k-300402" style="font-weight:bold;" target="_blank" title="Click to know more about 2K">2K</a>`<br/>`3k`</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> | |
| 37959. |
A circular frame made of 20 cm long thin wire is floating on the surface of water. If the surface tension of water is 70 dyne/cm. the required excess force to separate this frame from water will be : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/2800-1833959" style="font-weight:bold;" target="_blank" title="Click to know more about 2800">2800</a> dyne<br/>800 dyne<br/>280 dyne<br/>1800 dyne</p>Answer :A</body></html> | |
| 37960. |
Two bodies of equal masses moving with momentum of first double that of the second collide completely inelastically. If the total final kinetic energy is 9 joule, the initial kinetic energy of the body of greater momentum is: |
| 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> <a href="https://interviewquestions.tuteehub.com/tag/joule-526299" style="font-weight:bold;" target="_blank" title="Click to know more about JOULE">JOULE</a><br/>6 joule<br/>8 joule<br/>10 joule </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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 37961. |
A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ G = 6.67 xx10^(-1) SI unit and M= 6 xx 10^(24)kg ] |
| Answer» <html><body><p></p>Solution :`implies` Given , `r_p` = radius of perihelion = 2R <br/>`r_a` = radius of aphelion = 6R <br/> Hence , we can write, <br/> `r_(a) = a(1+e) = 6R "" ...(i)` <br/> `r_p = a (1-e) = 2 R ""...(ii)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/solving-1217196" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVING">SOLVING</a> equ (i) and (ii) ,we get <br/> <a href="https://interviewquestions.tuteehub.com/tag/eccentricity-964854" style="font-weight:bold;" target="_blank" title="Click to know more about ECCENTRICITY">ECCENTRICITY</a> , `e =1/2` <br/> Angular <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> remains unchanged <br/> `:. mv_pr_p = mv_ar_a` <br/> `:. (v_a)/(v_p) = r_p/r_a = (2R)/(6R) =1/3` <br/> Energy is same at perigee and apogee, <br/> `1/2mv_p^2-(GMm)/r_p=1/2mv_a^2-(GMm)/r_a` <br/>`:. v_p^2(1-1/9)=-2GM(1/r_a-1/r_p)=2GM(1/r_p-1/r_a)` <br/> (By putting `v_a=v_p/3`) <br/> `v_p=([2GM(1/r_p-1/r_a)]^(1/2))/([1-(v_a/v_p)^2]^(1/2))=[((2GM)/R(1/2-1/6))/((1-1/9))]^(1/2)` <br/> `=((3/2GM)/(8/9R))^(1/2)=sqrt(3/4(<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/R) ` = 6.85 km/s<br/> `v_p = 6.85km//s` <br/> For circular orbit of radius r, <br/> `v_c` = orbita velocity = `sqrt((GM)/r)` <br/> For `r = 6R ,v_c =sqrt((GM)/(6R)) = 3.23 km//s` <br/> Hence , to increase to a circular orbit at apogee, we have to increase the velocity by (3.22-2.28) = 0.95 km/s . This can be done by suitably firing rockets from the <a href="https://interviewquestions.tuteehub.com/tag/satellite-1195060" style="font-weight:bold;" target="_blank" title="Click to know more about SATELLITE">SATELLITE</a> .</body></html> | |
| 37962. |
The magnitude of potential energy per unit mass of the object at the surface of earth is .E.. Then escape velocity of the object is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(<a href="https://interviewquestions.tuteehub.com/tag/2e-300683" style="font-weight:bold;" target="_blank" title="Click to know more about 2E">2E</a>)` <br/>`4E^2`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sqrte-3057006" style="font-weight:bold;" target="_blank" title="Click to know more about SQRTE">SQRTE</a>`<br/>`sqrt(E//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> | |
| 37963. |
The motion of a particle is given by X=Asinomegat+B cos omegat the motion of the particle is |
| Answer» <html><body><p>not <a href="https://interviewquestions.tuteehub.com/tag/simple-1208262" style="font-weight:bold;" target="_blank" title="Click to know more about SIMPLE">SIMPLE</a> <a href="https://interviewquestions.tuteehub.com/tag/harmonic-1015999" style="font-weight:bold;" target="_blank" title="Click to know more about HARMONIC">HARMONIC</a> <br/>simple harmonicwith amplitude A+B<br/>Simple harmonic with amplitude `(A+b)/<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/>simple harmonic with amplitude `sqrt(A^(2)+B^(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> :D</body></html> | |
| 37964. |
A body of mass 5kg starts from the origin with an initial velocityof barU= (30i+40j)m//s. A constant force of F= (-hati - 5 hatj)N acts on the body . Find the time in which they- component of the velocity becomes zero . |
| Answer» <html><body><p></p>Solution :`baru= 30i + 40 j` ....... (1) <br/> ` u=u_(x) <a href="https://interviewquestions.tuteehub.com/tag/hati-2693493" style="font-weight:bold;" target="_blank" title="Click to know more about HATI">HATI</a> + u_(y) hatj`........ (2)<br/>`F= - hati - 5hatj` ........ (3) <br/> ` F= F_(x) hati + F_yhatj` ........ (4) <br/> comparing (1) and (2), (3) and (4) <br/> we have <br/> `u_y = 40 m//s , F_y =-5N` <br/> `F_y = ma_y , a_y = -1 m//s^(2)` <br/> `v_y = u_y + a_y xx 1 , <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> = 40 -1 xx t , t = 40` sec</body></html> | |
| 37965. |
The ends of a metal rod of length 40 cm are maintained at temperatures 100°C and 20°C respectively. At steady state condition, the temperature at a point at a distance 30 cm from the hot end of the rod is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</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>`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>^(@) C`<br/>`20^(@) C`<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>^(@) C`</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> | |
| 37966. |
A light metal stick of square cross-section (5cmxx5cm) and length 4m mass 2.5 kg ad is shown in the figure below determine its angle of inclination when the water surface is 1m above the hinge. What minimum depth of water above high will be required to bring the metal stick in vertical position. |
| 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> :`theta=30^(@)` <a href="https://interviewquestions.tuteehub.com/tag/depth-948801" style="font-weight:bold;" target="_blank" title="Click to know more about DEPTH">DEPTH</a> of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> `ge2m`</body></html> | |
| 37967. |
A horizontal uniform glass tube of 100cm length sealed at both ends contains 10cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively0°C and 80cm of mercury. If the air column at one end is kept at 0°C and the other end at 273°C, the pressure of air which is 0°C is (in cm of Hg) |
| Answer» <html><body><p>76<br/>88.2<br/>120<br/>132</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> | |
| 37968. |
A: A body is moving along a straight line such that its velocity varies with time as shown in figure. Magnitude of displacement of the body from t = 0 to t = 12s is the same as the distance travelled by it in the given time duration, R: For a unidirectional motion of a body, |displacement| = distance |
| Answer» <html><body><p>If both A and B 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 <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> is the correct explanation of A<br/>If both A and R are true but R is not correct <a href="https://interviewquestions.tuteehub.com/tag/explantion-2628196" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANTION">EXPLANTION</a> of A <br/>If A is true but R is false<br/>If A is false but R is true</p>Answer :A</body></html> | |
| 37969. |
For a body rolling along a level surface, without slipping the translational and rotational kinetic energies are in the ratio 2:1. The body is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/hollow-1028416" style="font-weight:bold;" target="_blank" title="Click to know more about HOLLOW">HOLLOW</a> sphere<br/>Solid cylinder<br/>Ring<br/>Solid sphere</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> | |
| 37970. |
A spring when compressed by 0.2 m develops a restoring force of 50 N. If a body of mass 10 kg is placed on it, calculate (i) the force constnat (I) of the spring (ii) the depression of the spring (iii) time period of oscillation when the weight is slightly displaced from equilibrium position. |
| Answer» <html><body><p></p>Solution :(i) The <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> constant `<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>=F/x=50/0.2=250Nm^(-1)` <br/> (ii) depression of the <a href="https://interviewquestions.tuteehub.com/tag/spring-11487" style="font-weight:bold;" target="_blank" title="Click to know more about SPRING">SPRING</a> `(mg)/k=(10xx9.8)/250=0.392m` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) Time period <br/> `T=2pisqrt(m/k)=2pisqrt(10/250)=(2xx3.14)/5=1.2568s`</body></html> | |
| 37971. |
The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 ms^(-1), what is the direction of the initial velocity. ? Take g = 9.8 ms^(-2). |
| 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> :`(<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>^(@))`</body></html> | |
| 37972. |
If vec(P)=hat(i)+2hat(j)-4hat(k)andvec(Q)=hat(i)+2hat(j)-hat(k) then (vec(P)+vec(Q))*(vec(P)-vec(Q)) is |
| Answer» <html><body><p>5<br/>15<br/>25<br/>115</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> | |
| 37973. |
One parsec is ......... |
| Answer» <html><body><p>`3.153 <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/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)` m<br/>`3.26 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)` m<br/>`30.84 xx 10^(15)` m<br/>`9.46 xx 10^(15)` m</p>Solution :`30.84 xx 10^(15)` m</body></html> | |
| 37974. |
What doesit mean when a height of a barometer is rising ? |
| Answer» <html><body><p></p>Solution :A rising height of a <a href="https://interviewquestions.tuteehub.com/tag/barometer-394088" style="font-weight:bold;" target="_blank" title="Click to know more about BAROMETER">BAROMETER</a> indicates increasing air <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> . It also indicates <a href="https://interviewquestions.tuteehub.com/tag/dry-960094" style="font-weight:bold;" target="_blank" title="Click to know more about DRY">DRY</a> <a href="https://interviewquestions.tuteehub.com/tag/atmosphere-13116" style="font-weight:bold;" target="_blank" title="Click to know more about ATMOSPHERE">ATMOSPHERE</a> (less humidity).</body></html> | |
| 37975. |
A machine gun has a mass of 20 kg . It fires 35 g bullets at the rate of 4bullets per second, with a speed of 400 ms^(-1). What force must be applied to the gun to keep it in position ? |
| 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> :Mass of <a href="https://interviewquestions.tuteehub.com/tag/gun-1014044" style="font-weight:bold;" target="_blank" title="Click to know more about GUN">GUN</a>, M=20 kg<br/> Velocity of bullet, `v=400 ms^(-1)`<br/> <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> taken to fire one bullet, `t=1/4` second <br/> If F be the average force required to hold the gun in position , then <br/> `Fxxt`= change in momentum of gun or `Fxxt = MV[`:.` initial momentum of gun is <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> ]<br/>`:. F=(MV)/t =(20xx0.7)/(1//4)N =56N`<br/>Aliter: Number of bullet <a href="https://interviewquestions.tuteehub.com/tag/fired-7262717" style="font-weight:bold;" target="_blank" title="Click to know more about FIRED">FIRED</a> per second, n=4 momentum given to the bullets per second = nmv<br/> Same momentum per second is gained by the gun . This gives the force acting on the gun and also the force applied to keep the gun in position .<br/> So , `F=nmv=4xx0.035xx400N=56N`</body></html> | |
| 37976. |
Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find the height from which they fall. (g - 10 ms^(-2)L - 80 caloric/g and I - 4.21/calorie) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/336-308095" style="font-weight:bold;" target="_blank" title="Click to know more about 336">336</a> m <br/> <a href="https://interviewquestions.tuteehub.com/tag/236-295263" style="font-weight:bold;" target="_blank" title="Click to know more about 236">236</a> m <br/> <a href="https://interviewquestions.tuteehub.com/tag/436-1876849" style="font-weight:bold;" target="_blank" title="Click to know more about 436">436</a> m<br/>536 m </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> | |
| 37977. |
A cyclist starts from the centre O of a circular park of radius 1km , reaches the edge P of the park , then cyclesalongthe circumference and returns to the centre along QO as shown in figure . If the round trip takes 10 min. what is the (a) net displacement . (b) average velocity and (c) average speed of the cyclist ? |
| Answer» <html><body><p></p>Solution :(a) Net displacement =zero as initial and final <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> of the cyclist is same <br/> (b) Average <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> `=("Net displacement")/("Total time taken")`=zero <br/> (c) Total distance travelled `=OP+PQ+QO` <br/> `=OP+1//4(2pir)+QO` <br/> `=(1+1+(1)/(4)xx2xx(22)/(7)xx1)`km <br/> `=(25)/(7)km` <br/>Average speed `=("Total path <a href="https://interviewquestions.tuteehub.com/tag/length-1071524" style="font-weight:bold;" target="_blank" title="Click to know more about LENGTH">LENGTH</a>")/("Total timetaken")` <br/> `=(25//7)/(10//60)km <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(-1)` <br/> `=21.43 km h^(-1)` .</body></html> | |
| 37978. |
A uniform sphere of radius a and density rho is divided in two parts by a plane at a distance b from its centre. Calculate the mutual attraction between two parts. |
| Answer» <html><body><p><br/></p>Answer :`1/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>^(2)rho^(2)<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>(a^(2)-b^(2))^(2)`</body></html> | |
| 37979. |
A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2), respectively. Its time period is……….. |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2pi-1838601" style="font-weight:bold;" target="_blank" title="Click to know more about 2PI">2PI</a> <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))`<br/>`2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))`<br/>`2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`<br/>`2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`</p>Solution :For <a href="https://interviewquestions.tuteehub.com/tag/shm-630759" style="font-weight:bold;" target="_blank" title="Click to know more about SHM">SHM</a> particle, <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>= omega sqrt(A^(2)-x^(2))` <br/> `therefore v_(1)= omega sqrt(A^(2)-x_(1)^(2))" and "v_(2)= omega sqrt((A^(2)-x_(2)^(2))` <br/> `therefore v_(1)^(2)= omega^(2)(A^(2)-x_(1)^(2))" and "v_(2)^(2)= omega^(2) (A^(2)-x_(2)^(2))` <br/> `therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]` <br/> `= omega^(2) [x_(2)^(2)-x_(2)^(2)]` <br/> `therefore omega^(2)= (v_(1)^(2)-v_(2)^(2))/((x_(2)^(2)-x_(1)^(2)))` <br/> `therefore omega = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` <br/> `therefore (2pi)/(T) = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` <br/> `therefore T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.</body></html> | |
| 37980. |
Standing waves can be poduced in the air column ofa pipe. Standing waves produced in an open pipe contain._____. |
| 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> :all <a href="https://interviewquestions.tuteehub.com/tag/harmonics-1016029" style="font-weight:bold;" target="_blank" title="Click to know more about HARMONICS">HARMONICS</a></body></html> | |
| 37981. |
Why do we slip on a rainy day ? |
| Answer» <html><body><p></p>Solution :On a rainy <a href="https://interviewquestions.tuteehub.com/tag/day-944776" style="font-weight:bold;" target="_blank" title="Click to know more about DAY">DAY</a>, the cofficient of friction between our feet and the wet ground gets much reduced. Consequently, the frictional <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a> `f_x^max=mu_sN` <a href="https://interviewquestions.tuteehub.com/tag/becomes-1994370" style="font-weight:bold;" target="_blank" title="Click to know more about BECOMES">BECOMES</a> very small. It <a href="https://interviewquestions.tuteehub.com/tag/may-557248" style="font-weight:bold;" target="_blank" title="Click to know more about MAY">MAY</a> cause us to slip.</body></html> | |
| 37982. |
A wire of length l, area of cross-section A and Young,s modulus of elasticity Y is suspended form the roof of a building. A block of mass m is attached at lower end of the wire. If the block is displaced from its mean position and released then the block starts oscillating. Time period of these oscillations will be |
| Answer» <html><body><p>`2pisqrt(Al)/(mY)`<br/>`2pisqrt(<a href="https://interviewquestions.tuteehub.com/tag/ay-386628" style="font-weight:bold;" target="_blank" title="Click to know more about AY">AY</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>)`<br/>`2pisqrt(ml)/(YA)`<br/>`2pisqrt(m)/(<a href="https://interviewquestions.tuteehub.com/tag/yal-2342566" style="font-weight:bold;" target="_blank" title="Click to know more about YAL">YAL</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> :C</body></html> | |
| 37983. |
When body is taken from the equator to the poles, its weight |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/remains-621920" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINS">REMAINS</a> same<br/>increases<br/><a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a><br/>increases at N-pole and decreases at S-pole</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> | |
| 37984. |
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^(3) at a temperature of 27 ^(@)C and 1 atm pressure. |
| 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> :`6.<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/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(<a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a>)`</body></html> | |
| 37985. |
A disc is at rest at the top of a rough incined plane. It rolls without slipping. At the bottom of inclined palne there is a vertical groove of radius .R.. In order to loop the groove, the minimum height of incline required is: |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/15r-276216" style="font-weight:bold;" target="_blank" title="Click to know more about 15R">15R</a>)/<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>`<br/>`(9R)/4`<br/>`(5R)/2`<br/>`(7R)/<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>`</p>Answer :A</body></html> | |
| 37986. |
A uniform rectangular plate is heated from 0^(@) to 100^(@)C. Initial area of plate is 10 cm^(2). The linear coefficient of the material of the plate is 18xx10^(-6)K. What is the shift of the centre of mass ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> cm<br/>2cm<br/>3cm</p>Answer :A</body></html> | |
| 37987. |
Show that a nonhomogeneous cylinder on a horizontal plane has a position of stable equilibrium when its centre of gravity is vertically below the axis O of the cylinder. |
| 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> :N//A</body></html> | |
| 37988. |
in the given progressive wave y = 5 sin (100 pi t -0.4 pi x) where yand x are in metre, t is in second. What is the (a) amplitude (b) wave length (c) frequency (d) wave velocity (e) particle velocity amplitude, |
| Answer» <html><body><p></p>Solution :Comparing give equation, <br/> `y = 5 <a href="https://interviewquestions.tuteehub.com/tag/sin-1208945" style="font-weight:bold;" target="_blank" title="Click to know more about SIN">SIN</a> (100 pi t 0.4 pix)` <br/> with `y = a sin (omega t - kx + phi)` we get <br/> `a= 5m, k = 0.4pi =- (2pi)/(lamda ) implies lamda = 5m` <br/> `omega = 100 pi = 2pi f implies f = <a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a> Hz` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> = (omega )/(k) = (100 pi)/(0.4pi) = <a href="https://interviewquestions.tuteehub.com/tag/250-297421" style="font-weight:bold;" target="_blank" title="Click to know more about 250">250</a> (m)/(s)` <br/> Maximum velocity of particle performing S.H.M. and taking part in the propagation of wave, <br/> `v _(max) = a omega` <br/> `= 5 xx 100 pi` <br/> `= <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> pi m//s`</body></html> | |
| 37989. |
Derive an expression for the gravitational potential energy of a body of mass 'm' raised to a height 'h' above the earth's surface. |
| 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> :The gravitational potential energy (U) at some height h is equal to the aount of <a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> required to take the object from the <a href="https://interviewquestions.tuteehub.com/tag/ground-1013213" style="font-weight:bold;" target="_blank" title="Click to know more about GROUND">GROUND</a> to that height h. <br/> `U=intvecF_(a),dvecr=underset(0)overset(h)int|vecF_(a)||dvecr|costheta` <br/> If the <a href="https://interviewquestions.tuteehub.com/tag/displacement-956081" style="font-weight:bold;" target="_blank" title="Click to know more about DISPLACEMENT">DISPLACEMENT</a> and the applied force are in the same upward direction then, the angle betweeen them is `theta=0^(@)`. Since, `cos0^(@)=1and|vecF_(a)|=mgand|dvecr|=dr` <br/> `U=mgunderset(0)oversethintdrimpliesmg[r]_(0)^(h)=<a href="https://interviewquestions.tuteehub.com/tag/mgh-1095458" style="font-weight:bold;" target="_blank" title="Click to know more about MGH">MGH</a>`</body></html> | |
| 37990. |
Assume that if the earth were made of lead of relative density 11.3, then what would be the value of acceleration due to gravity on the earth surface ? |
| Answer» <html><body><p></p>Solution :Since, <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> of earth can be given as <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rho-623364" style="font-weight:bold;" target="_blank" title="Click to know more about RHO">RHO</a>=` Relative density `xx` density of water `=11.3 xx 10^(3) "kgm"^(-3)` <br/> `:.` Acceleration due to <a href="https://interviewquestions.tuteehub.com/tag/gravity-18707" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITY">GRAVITY</a> on the earth's surface, <br/> `g=(<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/(R^(2))=(G)/(R^(2)).(4)/(3)piR^(3).rho` <br/> `=(4)/(3)xx(22)/(7)xx6.67xx10^(-11)xx6.4 xx10^(6)xx11.2xx10^(3)` <br/> `=22.21 <a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</a>^(-2)`</body></html> | |
| 37991. |
What is capillarity? Derive an expression for the ascent of liquid in a capillary. |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/rise-1189674" style="font-weight:bold;" target="_blank" title="Click to know more about RISE">RISE</a> or fall of a liquid in a narrow tube is called capillarity.<br/>Let us consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension.<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C07_E01_054_S01.png" width="80%"/> <br/>The surface tension force `F_(T)` acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, is resolved into two components.<br/>(i) Horizontal component T `sintheta` and <br/>(ii) Vertical component T `costheta` acting upwards, all along the whole circumference of the meniscus.<br/>Tota upward force `=(Tcostheta)(2pir)=2pirTcostheta` <br/>Where `<a href="https://interviewquestions.tuteehub.com/tag/theta-1412757" style="font-weight:bold;" target="_blank" title="Click to know more about THETA">THETA</a>` is the angle of contact, <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> is the <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> of the tube. Let `rho` be the density of water and h be the height to which the liquid rises inside the tube. Then,<br/>`{:(("the volume of"),("liquid column"),("in the tube,V")):}={:(("volume of the"),("liquid column of"),("radius r height h")):}+{:(("volume of the liquid of radius"),("r and height h-volume of"),("the hemisphere of radius of r")):}` <br/>`V=pir^(2)h+(pir^(2)xxr-(2)/(3)pir^(3))` <br/>`rArrV=pir^(2)h+(1)/(3)pir^(3)` <br/>The upward force supports the weight of the liquid column above the free surface, therefore,<br/>`2pirTcostheta=pir^(2)(h+(1)/(3)r)rhog` <br/>`rArr""T=((h+(1)/(3)r)rhog)/(2costheta)` <br/>If the capillary is a very <a href="https://interviewquestions.tuteehub.com/tag/fine-989069" style="font-weight:bold;" target="_blank" title="Click to know more about FINE">FINE</a> tube ofradius (i.e., radius is very small) then `(r)/(3)` can be neglected when it is compared to the height h. Therefore,<br/>`T=(rrhogh)/(2costheta)`</body></html> | |
| 37992. |
For one wire tied at both the ends with a tension of 450 N, linear density of mass is 0.05 g/cm. It sounds resonantly for two consecutive frequencies 420 Hz and 490 Hz. Find length of this wire. |
| Answer» <html><body><p></p>Solution :In the <a href="https://interviewquestions.tuteehub.com/tag/present-1163722" style="font-weight:bold;" target="_blank" title="Click to know more about PRESENT">PRESENT</a> <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>, <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> of `<a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>^(th)` harmonic <br/> `f _(n) = n f _(1) = 420 Hz` <br/> `f _(n +1) = (n +1) f _(1)= 490 Hz `<br/> Taking subtraction, `n f _(1) + f _(1) -nf _(1) = 490 -420` <br/> `therefore f _(1) = 70 Hz` <br/> `therefore (1)/(2L) sqrt ((T)/( mu ))=70`<br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C15_E02_051_S01.png" width="80%"/> <br/> `therefore L = (300)/(140) =2.1428 m ~~2.143 m`</body></html> | |
| 37993. |
Three identical vessels are filled to the same height with three different liquids A, B and C(rho_(A) gt rho_(B) gt rho_(C)). The pressure at the base will be |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> in all vessels<br/>Maximum in <a href="https://interviewquestions.tuteehub.com/tag/vessel-1445872" style="font-weight:bold;" target="_blank" title="Click to know more about VESSEL">VESSEL</a> A<br/>Maximum in vessel <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a><br/>Maximum in vessel</p>Answer :B</body></html> | |
| 37994. |
Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/geometrical-2093743" style="font-weight:bold;" target="_blank" title="Click to know more about GEOMETRICAL">GEOMETRICAL</a> centre of each. No, the CM may lie <a href="https://interviewquestions.tuteehub.com/tag/outside-1143075" style="font-weight:bold;" target="_blank" title="Click to know more about OUTSIDE">OUTSIDE</a> the body, as in case of a ring, a <a href="https://interviewquestions.tuteehub.com/tag/hollow-1028416" style="font-weight:bold;" target="_blank" title="Click to know more about HOLLOW">HOLLOW</a> sphere, ahollow cylinder, a hollow cube <a href="https://interviewquestions.tuteehub.com/tag/etc-975753" style="font-weight:bold;" target="_blank" title="Click to know more about ETC">ETC</a>.</body></html> | |
| 37995. |
A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg ? |
| Answer» <html><body><p></p>Solution :Height of the stairs <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 10 m <br/>Work done to burn 1 kg of fat = <a href="https://interviewquestions.tuteehub.com/tag/7000brbr-1917317" style="font-weight:bold;" target="_blank" title="Click to know more about 7000">7000</a> <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> cal <br/>`therefore` Work done to burn 5 kg of fat<br/> =5 x 7000 = 35000 k cal <br/> Work done towards burning of fat in up and down the stairs, <br/> `=mgh+1/2mgh` <br/> `=3/2mgh` <br/> `=3/2xx60xx10xx10` <br/> =9000 J <br/> `=9000xx1/4.2` cal <br/> =2142.85 cal <br/> `therefore` For energy of 2142.85 cal one trip (up and down) of stairs needed <br/> Hence , for energy of `35xx10^6` cal <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> trip (up and down) of stairs needed. <br/> `therefore n=(35xx10^6)/(2142.85)=0.01633xx10^6` <br/> =16330 <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a></body></html> | |
| 37996. |
If A=B+ (C)/(D+E)the dimensions of B and Care [M^(0)LT^(-1)] and [M^(0)LT^(0)], respectively. Find the dimensions of A, D and E. |
| Answer» <html><body><p>`[A]= [M^(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)L^(0)T^(-1)], [D]= [T], [E]= [LT]`<br/>`[A]= [<a href="https://interviewquestions.tuteehub.com/tag/mlt-2163963" style="font-weight:bold;" target="_blank" title="Click to know more about MLT">MLT</a>^(0)], [D]= [T^(2)], [E]= [T^(2)]`<br/>`[A]= [M^(0)LT^(-1)], [D]= [MT], [E]= [MT]`<br/>`[A]= [M^(0)LT^(-1)], [D]= [T], [E]= [T]`</p>Solution :As `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>+ (C)/(D+E)` <br/> `:. [D]= [E]` <br/> `:. [A]= [B]= [(C)/(D+E)]= [(C)/(D)]= [(C)/(E)] :. [A]= [B]= [M^(0)LT^(-1)]` <br/><br/> `[(C)/(D)]= [A]= [M^(0)LT^(-1) ]`<br/> `[D]= [E]= [(C)/(LT^(-1)] = ( [M^(0)LT^(0) ])/( [M^(0)LT^(-1) ])= [T]`</body></html> | |
| 37997. |
At what angle two vectors of magnitudes vecA +vecBand A - B act, so that their resultant is sqrt(3A^2 + B^2)? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a>^@` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a>^@` <br/>`60^@` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</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> :C</body></html> | |
| 37998. |
The moment of inertia of a wheel of radiu 20 cm is 40 kg m^(2). If a tangential force of 80 N applied on the wheel, its Rotational K.E. after 4s is |
| Answer» <html><body><p>16.2J<br/>51.2J<br/>25.6J<br/>24.8J</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> | |
| 37999. |
0.1m^3 of water at 80^@C is mixed with 0.3m^3 of water at 60^@C. The final temperature of the mixture is |
| Answer» <html><body><p>`65^@C`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/70-333629" style="font-weight:bold;" target="_blank" title="Click to know more about 70">70</a>^@C`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>^@C`<br/>`75^@C`</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> | |
| 38000. |
A particle has a time period of 1s under the action of a certain force and 2 s under the action of another force. Find the time period when the forces are acting in the same direction simultaneously. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> the period of oscillation be `T_1`under tre action of the force `F_1 and T_2`under the action of force `F_2`when acting <a href="https://interviewquestions.tuteehub.com/tag/separately-7264663" style="font-weight:bold;" target="_blank" title="Click to know more about SEPARATELY">SEPARATELY</a>. Lot F be the <a href="https://interviewquestions.tuteehub.com/tag/resultant-1187362" style="font-weight:bold;" target="_blank" title="Click to know more about RESULTANT">RESULTANT</a> force and the period of oscillation under the action of the resultant force <br/> `F=F_1+F_2` <br/> `momega^2y= momega_1^2y+momega_2^2y` <br/> `((2pi)/T)^2=((2pi)/T_1)^2+((2pi)/T_2)^2` <br/> `1/T^2=1/T_1^2+1/T_2^2` <br/> `T^2=(T_1^2T_2^2)/(T_1^2+T_2^2)` <br/> `T=(T_1T_2)/(<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(T_1^2+T_2^2))` <br/> `T_1=1 s, T_2 = <a href="https://interviewquestions.tuteehub.com/tag/2s-301125" style="font-weight:bold;" target="_blank" title="Click to know more about 2S">2S</a>` <br/> `T = (1xx2)/(sqrt(1^2+2^2))` <br/> = 0.894 s</body></html> | |