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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12851. |
A rod of mass 'm' and length l in placed on a smooth table. An another particle of same 'm' strikers the with velocity v_(0) in a distance perpendicular to rod at distance x(lt l//b) form its centre. Particle sticks to the rod Let omega be the angular speed of system after collision, then As x is increased from 0 to l//2, the angular speed omega : |
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Answer» Will continuosly INCREASE |
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| 12852. |
Three measurements of the time for 20 oscillations of a pendulum give t_(1)= 39.6 s, t_(2) = 39.9s and t_(3)= 39.5s The precision in the measurement, is |
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Answer» 0.1 sec |
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| 12853. |
The volume (V) of a monoatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is ____ |
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Answer» `2/7` In isobaric PROCESS `W=PDeltaV=muR DeltaT` ….(1) `DeltaQ=mu C_P DeltaT`…(2) For monatomic GAS MOLECULE f=3 `therefore C_P=f/2 R +R =5/2 R` …(3) `[because f=3]` `therefore W/(DeltaQ)=(muR DeltaT)/(mu C_P DeltaT)` [From equ. (1) and (2)] `therefore W/(DeltaQ)=R/C_P=R/(5/2R)` [From equ. (3)] `therefore W/(DeltaQ)=2/5` |
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| 12854. |
Two masses nm and m, start simultaneously from the intersection of twostraight lines with velocities v and nv respectively. It is observed that the path of their centre of mass is a straight line bisecting the angle between the given straight lines. Find the magnitude of the velocity of centre of mass. [Here theta=angle between the lines] |
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Answer» SOLUTION :`vecv_(cm)=(m_(1)vecv_(1)+m_(1)vecv_(2))/(m_(1)+m_(2))` `vecv_(cm)=(nmvhati+mnvcosthetahati+nmv sin theta HATJ)/(m+nm)` `v_(cm)=sqrt((nmv+mnvcostheta)^(2)+(nmvsintheta)^(2))/(m(1+n))` `v_(cm)=(nmvsqrt((1+costheta)^(2)+(SINTHETA)^(2)))/(m(1+n))` `=(nvsqrt(1+cos^(2)theta+2costheta+sin^(2)theta))/((1+n))` `=((nv)/(n+1))sqrt(2costheta/2)` `v_(cm)=(2nv"cos" theta/2)/(n+1)` |
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| 12855. |
An object A is dropped from rest the top of a 30m high building and at the same moment another object B is projected vertically upwards with an initial speed of 15 m/s from the base of the building. Mass of the object A is 2 kg while mass of the object B is 4 kg. Find the maximum height above the ground level attained by the centre of mass of the A and B system (take g=10m//s^(2) ) |
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Answer» Solution : `m_(1)=4kg, m_(2)=2KG` Initially 4 kg is on the ground `:.x_(1)=0` 2 kg is on top of the building `:.x_(2)=30m` `x_(CM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `=(0+2xx30)/(4+2)=10m` `:.` Initial height of CM =10m.  Initial velocity of CM, `u_(cm)=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))` `u_(cm)=(4xx15+0)/(4+2)=10m//s` upward. Acceleration of CM, `a_(cm)=g=10m//s^(2)` DOWNWARDS `:.` Maximum height attained by CM from initial position, `h_(cm)=(u_(cm)^(2))/(2g)=(10^(2))/(20)=5m` `:.`Maximum height attained by CM of 4 kg and 2 kg from the ground = 10 + 5 = 15 m |
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| 12856. |
Air expands from 5 litres to 10 litres at2 atm pressure. External workdone is |
| Answer» ANSWER :B | |
| 12857. |
A ball is thrown upward with an initial velocity of 100ms^(-1). After how much time will it return? Draw velocity -time graph for the ball and find the graph (i) the maximum height attained by the ball and (ii) height of the ball after 15s. Take g=10ms^(-2). |
Answer» Solution : Here, `u=100ms^(-1)""g=-10ms^(-2)` At HIGHEST point, v=0 As `v=u+gt therefore 0=100 -10 xx t` Time taken to reach highest point, `t=(100)/(10)=10S` The ball will return to the GROUND at t=20s Corresponding velocity -time graph of the ball is shown in (i) Maximum height attained the ball =Area of `triangleAOB=1/2 xx 10 xx 10=500m` (ii) Height attained after 15s= Area of `triangleAOB+"Area of "triangleBCD` `=500+1/2(15-10)xx (-50)=500-125=375m` |
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| 12858. |
In the previous question (i) work is done by the gas (ii) work is done on the gas (iii) heat is absorbed by the gas (iv) heat is given out by the gas |
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Answer» (i),(III) Work is done on the gas, heat is given out by the gas |
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| 12859. |
The average translational K.E of O_2 molecule at a particular temperature is 0.048 eV.The translational K.E of N_2 molecules in eV at the same temperature is |
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Answer» 0.0015 |
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| 12860. |
Calculate the rate of loss from a room through a glass window of area 1.5 m^(2) and thickness 2.5 xx 10^(-3) m, when the temperature of the room is 25^(@) C and that of the air outside is 10^(@) C. Assume that the inner glass surface is at the room temperature. [ Thermal conductivity of glass =1.2 "Wm"^(-1) "K"^(-1) ] |
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Answer» |
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| 12861. |
Seven identical rods of material of thermal conductivity k are connected as shown in Fig. All the rods are of identical length l and cross sectional area A If the one end B is kept at 100^@C and the other end is kept at 0^@C what would be the temperatures of the junctions C, D and E(theta_C,theta)D and theta_E) in the steady state? |
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Answer» `theta_Cgttheta_Egttheta_D`  This problem can be solved like electric current problem. `R_1`,`R_2`,`R_3`,`R_4`,`R_5`,`R_6` and `R_7` be the RATES of heat flow through AE,EB,AC,CD,CE,ED and DB, RESPECTIVELY. Since `R_1=R_2` `theta_E=50^@C` .(i) `R_5=R_6` `R_3=R_4+R_5=R_7`.(ii) `R_4+R_6=R_7` `(KA(theta_C-50))/(L)=(kA)/(l)(theta_C-50)+(kA)/(l)(theta_C-theta_D)=(kA)/(l)theta_D` `theta_C+theta_D=100` `2theta_C-2theta_D=50impliestheta_C=62.5^@C` `theta_D=37.5^@C` `theta_Cgttheta_Egttheta_D` |
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| 12862. |
Two equal, straight strips of two different metals are fastened together parallel to each other, a small.fixed distanced apart to form a bimetallic strip. Find the radius of curvature of the bimetallic strip when it is heated from 0^(0)C to t^(0) C. |
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Answer» Solution :Let A and B be two straight strips fastened parallel to each other at `0^(0)`C . When heated to `t^(0)C` the strips of different metals expand differently. So, the BIMETALLIC STRIP BENDS. The radius of curvature of strip A is `r_(1)` and that of B is `r_(2)`. If `l_(0)` is the original length of the metal strips at `0^(0)C and l_(1) , l_(2)` are the LENGTHS of strips A , B respectively at `t^(0)C ,` then `l_(1) = l_(0) (1 + alpha_(1) t)` and `l_(2) = l_(0) (1 + alpha_(2) t)` .... (i) Where `alpha_(1) and alpha_(2)` are the coefficient of linear expansion of strips A and B respectively. their common centre of curvature O and `r_(1),r_(2)` are the radii of curvature of strips A, B respectively, then `l_(1) = r_(1) phi and l_(2) = r_(2) phi` ............ (ii) ` (##AKS_NEO_CAO_PHY_XI_V01_PMH_C12_SLV_021_S01.png" width="80%"> Substituting `l_(1) and l_(2) ` from equation (i) `l_(0) (1 + alpha_(1) t) = r_(1) phi and l_(0) (1 + alpha t) = r_(2) phi)` `(r_(1) - r_(2) ) phi = l_(0) (alpha_(1) - alpha_(2) )t ` `phi = (l_(0) (alpha_(1) - alpha_(2))t)/((r_(1) - r_(2))) = (l_(0) (alpha_(1) alpha_(2))t)/(d) = (therefore r_(1) - r_(2) = d)` `l_(0) ( 1 + alpha_(1) t) = r_(1) phi and l_(0) ( 1 + alpha_(2) t ) = r_(2) phi` ` (##AKS_NEO_CAO_PHY_XI_V01_PMH_C12_SLV_021_S02.png" width="80%"> Adding `l_(0) (1 + alpha_(1) t + l + alpha_(2) t) = (r_(1) + r_(2) ) phi` `l_(0) (2 + (alpha_(1) + alpha_(2))t) = (r_(1) + r_(2)) (l_(0) (alpha_(1) - alpha_(2))t)/(d)` ` 2 = (r_(1) + r_(2)) ((alpha_(1) - alpha_(2))t)/(d)` `because (alpha_(1) + alpha_(2))`t is neglected when compared with 2 ) `(r_(1) + r_(2))/(2) = (d)/((alpha_(1) - alpha_(2))t)` `(r_(1) + r_(2))/(2) ` is the average radius of curvature of the bimetallic strip when heated. It is denoted by r. `r = (d)/((alpha_(1) - alpha_(2))t)` When the bimetallic strip is cooled by `1^(0)` C , it bends in opposite direction with same average radius of curvature. |
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| 12863. |
In the diagram shown in figure, match the following columns (take g=10 ms^(-2)) {:("Column I","Column II"),("(A) Normal reaction","(p) 12 SI unit"),("(B) Force of friction","(q) 20 SI unit"),("(C) Acceleration of block","(r) zero"),(,"(s) 2 SI unit"):} |
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Answer» `F mu_(k)N=12N` Since, `20sqrt(2) cos 45^(@) gt muN` block will MOVE and kinetic friction will act, `a=(20sqrt(2) cos 45^(@)-mu_(k)N)/(m)=(20-12)/(4)=2 MS^(-2)` |
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| 12864. |
Find vecAxxvecB if vecA=hati-2hatj+4hatk"and"vecB=3hatj+2hatk |
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Answer» SOLUTION :`vecAxxvecB=|(veci,vecj,veck),(1,-2,4),(0,3,2)|` `=HATI(-4-(12))-hatj(2)+K(3))=-16hati-2hatj+3hatk` |
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| 12865. |
An air bubble released at the bottom of a lake, rises and in reaching the top, its radius is found to be doubled. If the atmospheric pressure is equivalent to H metre of water column, find the depth of the lake (Assume that the temperature of water in the lake is uniform) |
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Answer» Solution :Volume of the air bubble at the bottom of the lake `(V_(1))=4/3pir^(3)` Volume of the air bubble at the surface of the lake `(V_(2))=4/3pi(2r)^(3)` PRESSURE at the surface of the lake `(P_(2))=H` metre of water COLUMN. If .h. is the DEPTH of the lake, the pressure at the bottom of the lake `(P_(1))=(H+h)` metre of water column. Since the temperature of the lake is uniform, ACCORDING to Boyle.s law, `P_(1)V_(1)=P_(2)V_(2)` `(H+h)(4/3pir^(3))=H[4/3pi(2r)^(3)], (H+h)=8H, h=7H` |
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| 12866. |
A pond has an ice layer of thickness 3 cm. If K of ice is 0.005 CGS units, surface temperature of surroundings is -20^(0)C, density of ice is 0.9 gm/cc, the time taken for the thickness to increase by 1 cm is |
| Answer» Answer :C | |
| 12867. |
Consider P-V diagram for an ideal gas shown in Fig 12.2. Out of the following diagrams which represents the T-P diagram ? |
| Answer» ANSWER :C | |
| 12868. |
A pick up van negotiates an unbanked curve of radius 40 m with a crate of eggs in the middle of its flat bed. If the coefficient of static friction between the crate and the truck, is 0.55 what is the maximum speed the van can take, so that the crate does not slide ? |
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Answer» Solution :`r=40 m, MU = 0.55` Maximum speed = `v_("max") = ?` `v_("max") = sqrt(murg) = sqrt(0.55 XX 40 xx 9.8) = 14.68` m/s |
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| 12869. |
PQR is a rigid equilateral triangular frame of side length 'L'. Forces F_(1), F_(2) and F_(3) are acting along PQ, QR and PR. Find the relation between the forces if system is in equilibrium. |
Answer» Solution :If the SYSTEM is in rotational equilibrium find thhe relation between the forces.  Perpendicular DISTANCE of any force shown in the figure from centroid .C. of triangle is `(L)/(2sqrt(3))` The forces `F_(1)` and `F_(2)` produce anti-clockwise turning EFFECT where as `F_(3)` clockwise turning effect about .C.. Since the system is in rotational equilibrium the total torque acting on the system about the centroid is zero. `:.F_(1)xx(L)/(2sqrt(3))+F_(2)xx(L)/(2sqrt(3))F_(3)xx(L)/(2sqrt(3))=0` Hence `F_(1)+F_(2)-F_(3)=0` `:.F_(3)=F_(1)+F_(2)` |
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| 12870. |
The initial state of certain gas (P_(i)V_(i)T_(i)).It undergoes expansion till its volume becomes V_(f) at constant temperature T. The correct plot of P-V diagram for it is |
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Answer»
At CONSTANT temperature `P_(i)V_(i)=nRT and P_(f)V_(f)=nRT rArr P_(i)V_(i) = P_(f)V_(f)` In the given P-V diagram (a) at every point on the plot P-V is constant which SHOWS a process at constant temperature i.e ISOTHERMAL process. |
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| 12871. |
A disc is rotating at an angular velocity omega_(0). A constant retardation torque is applied on it to stop the disc. After a certain time at which some number of rotation of the disc have been performed so that total angle rotated is theta_(1) and that only 2/3 th of these ratios will further stop the disc. find the retarding force. |
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Answer» `(11 omega_(0)^(2))/(14 theta_(1))` `n_(1)=(theta_(1)//2pi)` `n_(2)=2/3n_(1)=2/3((theta_(1))/(2pi))=(2/5 theta_(1)).1/(2pi)` `omega_(0)^(2)=omega_(0)^(2)-2alphatheta_(1).....(1)` and `0=omega_(1)^(2)-2alpha(2/5 theta_(1)).....(2)` From eqns. `(1)` and `(2)` `(4alphatheta_(1))/5=omega_(0)^(2)-2alphatheta_(1), alpha=(5 omega_(0)^(2))/(14 theta_(1))` |
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| 12872. |
Calculate the height to which water will rise in capillary tube of 1.5 mm diameter. Surface tension of water is 7.4 xx 10^(-3) Nm^(-1). |
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Answer» Solution :Here, `r = (1.5)2 = 0.75 mm = 0.75 XX 10^(-3) m` `sigma = 7.4 xx 10^(-3) Nm^(-1)` For water, `rho = 10^3 m^(-3)` ANGLE of CONTACT`"" theta = 0^@` `h = (2 sigma cos theta)/(r rho G) = (2 xx 7.4 xx 10^(-3) xx cos 0^@)/(0.75 xx 10^(-3) xx 10^3 xx 9.8)` `h = 0.002014 m` |
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| 12873. |
A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed T, the thickness of the shell not be less than. |
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Answer» `(4pi KTR^(2))/(P)` |
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| 12874. |
Pendulum beating seconds at the equator (g= 978 cm // "sec"^(2) ) is taken to Antarctica(g = 983 cm/sec), its length is to be increased by |
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Answer» 0.1 cm |
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| 12875. |
An air bubble of radius 'r' is formed at a depth 'h' below the surface of water of density 'd' and if 'p_(0)' is the atm pressure, T is surface tension of water, then the pressure inside the bubble is |
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Answer» <P>`p_(0)+(2T)/(R )` |
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| 12876. |
"We slip on a muddy road". Why ? |
| Answer» Solution :The pressure of MUD decreases friction. The WATER on the muddy ROADS acts as a thin layer in between our FEET and the road, thereby breaking the interlocking. This reducesfriction . So we slip. | |
| 12877. |
A ship A is moving Westwards with a speed of 10 km h^(-1) and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h^(1). The time after which the distance between them becomes shortest is |
| Answer» Answer :D | |
| 12878. |
A black body having an area of 2xx10^(-4)m^(2) for its radiating surface radiates energy of 16.42 J in 15 minutes.What is the temperature of the body ? |
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Answer» |
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| 12879. |
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? |
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Answer» Solution :SINCE stick is in rotational equilibrium, the TOTAL torque of all the FORCES about the resultant .R. is zero. Taking the turning effects about the point of action of the resultant .R. we have `10 g xx 3=mg xx 5` on solving m = 66 g 
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| 12880. |
The system shown in figure consists of three springs and two rods as shown. These springs are initially relaxed and there is no friction. The temperature of each of the rod is increased by DeltaT. The coefficient of linear expansion of the material of rods is equal to alpha. |
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Answer» The energy STORED in spring `S_` is `(81)/( 242) KL^(2) alpha (Delta T)^(2) ` |
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| 12881. |
The potential energy of a particle (U_x)executing SHM is given by |
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Answer» `U_x = K/2 (X - a)^2` |
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| 12882. |
A circular disk of moment of inertia I_(t) is rotating in a horizontal plane , about its symmetry axis , with a constant angular speed omega_(i) . Another disk of moment of inertia I_(b) is dropped coaxially onto the rotating disk . Initially the second disk has zero angular speed . eventually both the disks rotate with a constant angular speed omega_f . The energy lost by the initially rotating disc to friction is |
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Answer» `(1)/(2) (I_(B)^(2))/((I_(t) + I_(b))) omega_(i)^(2)` |
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| 12883. |
Find the temperature of 149^(@) F on kelvin scale. |
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Answer» SOLUTION :`(F-32)/(180)=(T-273)/(100)` `(149-32)/(180)=(T-273)/(100)IMPLIES(117)/(9)=T-273` T=286 k |
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| 12884. |
The M.I. of a uniform circular disc about a diameter is I. Its M.I. about an axis perpencular to its plane and passing through a point on its rim will be |
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Answer» 2I |
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| 12885. |
Length of a string tied to twop rigid supports is 40cm. maximum length of a stationary wave produced on it is |
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Answer» 20 cm |
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| 12886. |
A body of mass 0.5 kg travels in a straight line with velocity v = ax^(3/2)where a = 5 m^(-1/2) s^(-1) . The work done by the net force during its displacement from x = 0 to x = 2 m is |
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Answer» `1.5` J `m=0.5` kg `a =5 m^(-1/2) s^(-1)` Now `a_(0)=(dv)/(dt)` ` =v(dv)/(dx)` `=ax^(3/2) d/(dx)(ax^(3/2))` `=ax^(3/2) XX axx 3/2 XXX^(1/2)` `=3/2 a^(2)x^(2)` `= ax^(3/2)xxaxx 3/2 xxx^(1/2)` ` = 3/2a^(2)x^(2)` Now , `ma_(0)=m 3/2a^(2)x^(2)` Force , F SMALLER work done dW =Fdx Work done = `int_(xto0)^(x=2) F dx` ` = int_(0)^(2) 3/2 ma^(2)x^(2)dx ` `=3/2ma^(2)xx((x^(3))/3)_(0)^(2)` `=1/2ma^(2)xx8` `=1/2xx(0.5)xx(25)xx8 = 50 J ` |
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| 12887. |
A raindrop of mass 1.00 g falling from a height of 1m hits the ground with a speed of 50ms^(-1)Calculate (a) the lossof PE of the drop (b) the gain in KE of the drop ( c ) Is the gain in KE equal to loss of PE ? If not why ? Take, g =10 ms^(-2) |
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Answer» Solution :m = 1.00 g ` = 1xx10^(-3) kg` h = 1 km = `10^(3)` m ` g = 10 m//s^(2)` `V = 50 m//s ` (a) Loss in PE = mgh `= 1xx 10^(-3) xx10 xx10^(3)` = 10 J (b) Gain in KE ` = 1/2 MV^(2)` ` 1/2 xx1xx10^(-3) xx(50)^(2)` `=1/2 xx10^(-3) xx2500` `= 1.250 J ` No , gain in KE is not equal to the loss in its PE , because a aprt of PE is used in doing work againstthe viscous DRAG of air . |
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| 12888. |
A ball rolls down an inclined track 2 m long 4s. Find (i) acceleration (ii) time taken to cover the second metre of the track and (iii) speed of the vall at the bottom of the track. |
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| 12889. |
Whichof the following relations are correct for any type of motion ? (A) vecv_(av)=1/2[vecv(t_1)+vecv(t_2)] (B) vecv_(av)=(vecr(t_2)-(t_1))/((t_2-t_1)) (C ) vecv(t)=vecv(0)+vecat (D) vecr(t)=vecr(0)+vecv(0)*t+1/2vecat^2 (E) veca_(av)=(vecv(t_2)-vecv(t_1))/((t_2-t_1)) All the symbols have usual meaning. The subscript 'av' has been usedto denote the average of a quantity in the time interval t_1 to t_2. |
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| 12890. |
Three identical balls each of mass 5 kg are connected with each other as shown in figure, and rests over a smooth horizontal table. At moment t = 0 ball B is given velocity 9 m/sec then velocity of A in direction of b just before collision in m/s is ________ the velocity with which A collides with C in m/s is _______ |
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Answer» |
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| 12891. |
Heat is supplied to a certain homogeneous sample of matter.at a uniform rate. Its temperature is plotted against time, as shown. Which of the following conclusions can be drawn ? |
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Answer» Its SPECIFIC heat capacity is greater in THESOLID STATE than n the liquid state |
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| 12892. |
The fundamental frequency in an open organ pipe is equal tothe third harmonic of a closed organ pipe . If the length of the closed organ pipe is 20 cm , the length of the open organ pipe is |
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Answer» 12 . 5 cm `n_(2) = (3V)/(4l) ` [ l = length of closed organ pipe V = velocity of sound] Fundamental frequency of open organ pipe, `n_(0) = (V)/(2l^(.)` [`l^(.) =` length of open organ pipe] `:.(3V)/(4l)=(V)/(2l^(.))or, l^(.) = (4l)/(3xx2) = (2l)/(3) = (2xx20)/(3)` or,`l^(.) = 13 . 33` cm |
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| 12893. |
A bock A of mass 6 kg is applied on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 4kg at the other end. Find the acceleration of the system and tension in the thread (given g=10 ms^(-2)) |
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Answer» `2m//s^(2), 12N` |
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| 12894. |
Vector vecA is 2cm long and is 600 above the x-axis in the first quadrant, vector vecB is 2cm long and is 600 below the x-axis in the fourth quadrant. Find vecA+vecB. |
Answer» Solution : `VECR=vecA+vecB` `vecR = 2cos60^(0)hat(i)+2sin60hat(J)+2cos60hat(i)-2sin60hat(j)` `vecR = 4cos60hat(i) = 2hat(i)` `:.2cm, ALONG x-axis |
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| 12895. |
A flat plate moves normally with a speed v_(1) towards a horizontal jet of water of uniform area of cross section. The jet discharges water at the rate of volume V per second at a speed of v_(2). The density of water is rho. Assume that water splashes along the surface of the plate at right angles to the original motion. The magnitude of the force acting on the plate due to the jet is |
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Answer» SOLUTION :`"Force acting on the PLATE F"=(DP)/(dt)=u_(r)=(dm)/(dt)` `"Since, "Av_(2)=V rArr (dm)/(dt)=A(v_(1)+v_(2))rho=(V)/(v_(2))(v_(1)+v_(2))rho` `(u_(r)=v_(1)+v_(2)="VELOCITY of water COMING out of jet w.r.t plate")` `F=(v_(1)+v_(2)).(V)/(v_(2))(v_(1)+v_(2))rho=(V)/(v_(2))rho=(V)/(v_(2))(v_(1)+v_(2))^(2)rhoN` |
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| 12896. |
At a pressure of 3 atm air (treated as an ideal gas) is pumped into the tubes of a cycle rickwise. The volume of each tube at given pressure is 0.004 m^(3). One of the tubes gets punctured and the volume of the tube reduces to 0.0008 m^(3). Find the number of moles of air that have leaked out? Assume that the temperature remains constant at 300K. (R = 25//3J mol^(-1)K^(-1)) |
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| 12897. |
At what angle do the two forces (P+Q) and (P-Q) act so that the resultant is sqrt(3P^(2)+Q^(2))? |
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Answer» SOLUTION :Use `R=sqrt(A^(2)+B^(2)+2ABcostheta)` `R=sqrt(3P^(2)+Q^(2)),A=P+Q,B=P-Q` Solve, `theta=60^(@)` |
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| 12898. |
The height of the dam, in an hydroelectric power station is 10 m. In order to generate 1 MW of electric power, the mass of water (in kg) that must fall per second on the blades of the turbine is |
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Answer» `10^6` As `mgh =` work DONE `:. (dm)/(dt) gh = `POWER Given `h = 10 m` `:. (dm)/(dt) = (1 XX 10^6 W)/(10 MS^(-2) xx 10 m) = 10^4 kg s^(-1).` |
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| 12899. |
A particle is moving in a circle of radius R in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is Vo, the time taken to complete the first revolution is |
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Answer» `(R )/(v_0)` TANGENTIAL acceleration,`a_t = (dv)/(dt)` As` a_t-a_n `( given ) ` therefore(dv )/( dt ) = (v^2)/(R )or (dt)/(R )= (dv )/(v^2)` Integrating the above EQUATION, we get `int_(0)^(1)(dt)/(R )= int_(v_0)^(v)(dv )/(v^2) or(t)/(R )=- [ (1)/(v)]_(v_0)^(v) orv= (v_0R )/( (R-v_0 t))` As `v=(dr)/(dt) = (v_0 R )/((R-v_0t)) thereforetherefore(dr )/(R )= (v_0dt)/( (R-v_0 t))` INTEGRATE the above equation, we get` int_(0)^(2pi R )(dr )/(R )=int_(0)^(T)(v_0 dt)/(R-v0 t))` on simplification, we get `T= (R )/(v_0)(1-e^(-2pi ))` |
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| 12900. |
Calculate the rms velocity of oxygen molecules at NTP,the molecular weight of oxygen at being 32. |
| Answer» SOLUTION :416.2 `ms^-1` | |
