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InterviewSolution
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.
| 301. |
(a) It is known that density `rho` of air decrease with height y(in metres ) as `rho =rho_(0)e^(-y//y_(0))` where `rho_(0)=1.25kgm^(-3)` is the density at see level , and `y_(0)` is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmospere remians a constant (isothermal conditions). Also assume that the value of `g` remains constant. (b) A large He balloon of volume `1425m^(3)` is used to lift a payload of `400g`. Assume that the balloon maintains constant radius as it rises. How high does it rise? [take `y_(0)=800m` and `rho_(He) = 0.18kg//m^(3)`]. |
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Answer» We know that the rate of decrease of density `rho` of air with height y is directly proportional to density `rho` i.e., `-(d rho)/(dy) prop rho or (dp)/(dy) =-krho` where `K` is a constant of proportional . Here -ve sign shows that `rho` decreases as y increase. `:. (d rho)/(rho)=-K dy` Intergrating it within constions , as `y` changes from `0` to `y` , density change from `rho_(0)` to `rho`, we have `int_(rho_0)^(rho) (d rho)/(rho) = -int_(0)^(y) K dy` `[ log_(e)rho]_(rho_0)^(rho) = -Ky` or `log_(e)rho-log_(e)rho_(0)=-Ky` or `log(rho)/(rho_0) = -Ky` or `(rho)/(rho_0) = e^(-Ky)` or `rho=rho_(0)e^(-Ky)` Here `K` is a constant . suppose `y_(0)` is a constant such that `K=1//y_(0)`, then `rho=rho_(0)e^(-y//y_(0))`. (b) The balloon will rise to a height, where its density becomes equal to the air at that height. Density of balloon, `rho = (mass)/(volume) = ("pay load" + "mass of He")/("volume") = ((400+1425xx0.18)kg)/(1425m^(3)) = (656.5)/(1425) kg//m^(3)` As, `rho = rho_(0)e^(-y//y_(0)) :. (656.5)/(1425) = 1.25e^(-y//8000)` or,`e^(-y//8000) = (656.5)/(1425xx1.25) or e^(y//8000) = (1425xx1.25)/(656.5) = 2.7132` Taking log on both the sides `(y)/(8000) = log_(e)2.7132=2.3026 log_(10)2.7132 = 2.3026xx0.4335 ~~1` `y=8000xx1 = 8000m = 8km`. |
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| 302. |
A cylinderical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range h is equal toA. `H//4`B. `H//2`C. `3H//4`D. `H` |
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Answer» Correct Answer - B Horizontal velocity of water out of hole, `u=sqrt(2 g h)` Height of hole from ground level =(H - h). The time taken buy water to cover veertical distance (H - h) will be `(H - h)=(1)/(2)g r^(2)` or `t= sqrt(2(H - h)//g)` `:.` Horizontal range, `R=u t =sqrt(2 g h)xx sqrt(2(H - h)//g)` `=2sqrt(h(H - h))` Horizontal range will be maximum if `dR//dh =0` i.e., `2xx(1)/(2) (hH - h^(2))^(-1//2)xx(H-2 h)=0` or `H=2 h` or `h=H//2` |
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| 303. |
A small hollow sphere which has a small hole in it is immersed in water to a depth of 40 cm, before any water is penetrated into it. If the surface tensionof water si `0.073 Nm^(-1)`, find the radius of the hole. |
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Answer» Correct Answer - `3.7 xx 10^(-5)m` `P = 2S//r` or `r = 2 S//p` `= 2xx0.073//(0.40xx 10^3 xx9.8) = 3.7 xx 10^(-5)m` |
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| 304. |
A wooden ball of density D is immersed in water of density d to a depth h//2 below the surface of water and then relased. To what height will the ball jump out of water ? |
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Answer» Correct Answer - `((d)/(D) - 1) (h)/(2)` Let V be the volume of the ball Effective upward thrust on ball = V dg - V D g Upward ac celeration of ball, `a = (V d g- VD g)/(VD) = ((d - D)/(D))g` Velocity of the ball on reaching the surface `upsilon = sqrt(2 ah//2) = sqrt(ah)` If the ball rises through height H outside the liquid, the then `upsilon = sqrt(2 g H)` `:. sqrt(ah) = sqrt(2 g H)` or `H = (ah)/(2g) = ((d -D)/(D)) (gh)/(2g) = ((d)/(D) - 1) (h)/(2)` |
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| 305. |
A wooden ball of density ` rho` is immersed in a liquid of density `sigma` to a depth H and then released. The height h above the surface of which the ball rises will beA. HB. `(sigma)/(rho)`C. `((sigma-rho)/(rho))H`D. `(rho)/(sigma)H` |
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Answer» Correct Answer - C Let the volume of wooden ball be V. Weight of the ball `=V rho g`. Upward thrust on the ball due to the liquid `=V sigma g`. Net upward force on the ball `=V sigma g- V rho g= V g(sigma-rho)` Upward ac celeration in the ball `a=("force")/(mass)=(V g(sigma-rho))/(V rho)=((sigma-rho)g)/(rho)` Velocity on reaching the surface `v=sqrt(2 a H)` , If h is the height to which the ball rises above the surface of liquid, then `v=sqrt(2 gh)` . Hence `sqrt(2 a H)=sqrt(2 gh)` or `2a H=2 gh` or `h=(a H)/(g)=((sigma-rho)/(rho))(g H)/(g)= ((sigma-rho)/(rho))H` |
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| 306. |
If a small ping pong ball is placed in a vertical jet of water or air, it will rise to a certain geight above the nozzle and will be spinning there. |
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Answer» If a small ping pong ball is placed in a vertical jet of water the high velocity of water jet takes the ball upwards along with it through a certain height. Due to high velocity of jet of water, the pressure between the ball and the jet of water decrease i.e. becomes less than atmospheric pressure, therefore , this pressure difference pushes the ball against the jet and the ball remains suspended. Now jet of water exerts a tangential force on the surface of ball. Due to it, the ball spins. |
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| 307. |
How does the knowlefge of elasticity help in the construstion of steel bridges and sports domes, which remain safe during stroms and quakes. |
| Answer» While constructing steel bridges or sports domes, the triangular shaped structure of steel is used as it does not twist and collapse, whenever it is under pressure due to storms or quakes. | |
| 308. |
We would like to perpare a scale whose length does not change with temperature. It is proposed to prepare a unit scale f this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length (both components) would change in such a way that differnece between theri lenght rermain constant. If `alpha_(iron) = 1.2 xx 10^(-5)//K and alpha_(brass) = 1.8 xx 10^(-5)//K,` what should we take as lenght of each strip ? |
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Answer» Let `l_i and l_b` be the length of iron scale and brass at temper4ature `Delta theta .^@C.` Let `l_(io)` and `l_(bo)` be the legnth of iron scale and brass scale at temperature `0^@C.` Then as per question, `l_(omega) - l_(bo) = 10 cm =l_i - l_b` ...... (i) `l_i = l_(io)(1 + alpha_i Delta theta)` `l_b = l_(bo) (1 +alpha_b Delta theta)` `:. l_i - l_b = (l_(io) - l_(bo) + Delta theta (I_(io) alpha_i - I_(bo) alpha b)` Since `I_i - I_b = I_(io) - I_(bo) = `a constant as per question hence `I_(io) alpha_i - I_(bo) alpha_b = 0` or `I_(io) alpha_i = I_(bo) alpha_b` or `(I_(io))/(I_(bo)) = (alpha_b)/(alph_i) = (1.8 xx 10^(-5))/(1.2 xx 10^(-5)) = (3)/(2)` ..... (ii) As per question `(I_(io))/(I_(bo)) = (3)/(2)` or `I_(io) = (3)/(2) I_(bo)` Putting this value in (i), we have `(3)/(2)I_(bo) - I_(bo) = 10` or `I_(bo) = 20cm` and `I_(io) = (3)/(2) xx 20 = 30 cm` |
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| 309. |
100 g of water is supercooled to - `10^@C.` At this point, due to same disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperautre of the resultant mixture and how much mass would freeze ?`[s_w = 1 cal//g// .^@ C and L_(Fusion)^(w) = 80 cal//g]` |
| Answer» Heat required to bring 100 g super cooled water from `- 10^@`C to `0^@C` where ice is formed, ` = ms Delta T =100 xx 1 xx[0 - (-10)] = 1000 cal.` Thus the temperature of the resultant mixture (i.e. water and ice) `= 0^@C` If m gram of water freezes into ice. then `m xx 80 = 1000 cal` or `m = (1000)/(80) g = 12.5 g` | |
| 310. |
A solid shell loses half of its weight in water. Relative density of shell is 5.0 What fraction of its volume is hollow ? |
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Answer» Correct Answer - `3//5` Let x freaction of its volume be hollow. Then weight of the shell ` = (V - xV) xx(5.0 xx 10^3)xxg.` Weight of water displacedc `= V xx10^3 g` Loss in weight of shell = weight of water dispalced `:. (1)/(2) (V - xV) xx 5 xx10^3 xxg = Vxx 10^3 xx g` On solving `x = 3//5` |
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| 311. |
There are two electric poles, one is with hollow structure and other with a solid structure. They are made from the same and equal amount of material. Which electric pole do you perfer? Why. |
| Answer» We perfer hollows structure electorn pole because hollow structural electric pole. | |
| 312. |
A liquid of density `rho` is filled in a beaker of cross section A to a height H and then a cylinder of mass M and cross-section a is made to float in it as shown in If the atmospheric pressure is `P_0` find the pressure (a) at the top face B of the cylinder (b) at the bottom face C of the cylinder and (c ) at the base D of the beaker. (d) Can ever these pressure be equal ? |
| Answer» (a) As the top face B of cylinder is in contact with atmospherice air, so the pressure at B is atmospheric pressure i.e. `P_B = P_0` (b) The pressure at C is due to atmospherice air and also de to weight of the cylinder, i.e., ` P_C = P_0 +(Mg)/(A)` (c ) The pressure at D is due to atmospheric air, weight of cylinder and weight of liquid in cylinder i.e., `P_D = P_0 + ((Mg + H rho g A))/(A) = P_0 + (Mg)/(A) + H rho g` (d) in the absence of gravity i.e., the system is in free fall (as is in satellite), then g =0 in this case `P_B = P_C = P_D = P_0.` | |
| 313. |
A beaker filled with water at `4 .^(@)C` over flows if the temperature of water increases or decreases. Explain why? |
| Answer» The water has anamalous expansion. Its density is maximum at `4^(@)C` and its volume is minimum at `4^(@)C`. When temperature of water increases or decreases from `4^(@)C` it expands. Its volume increases so the beaker filled with water over flows. | |
| 314. |
Assertion : The angle of contact of a lilquid with a solid decreases with increase in temperature. Reason : With increase in temperature, the surface tension of the liquid increases.A. both Assertion and Reason are true and the Reason in the correct explanation of the Assertion.B. both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. Assertion is true but the Reason is false.D. both Assertion and Reason are false. |
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Answer» Correct Answer - D Here both, assertion and reason are false. Angle of contact increases with increase in temperature. Surface tension decreases on increasing temperature. |
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| 315. |
Mercury in a capillary tube suffers a depression of 13.2 mm. Find the diameter of the tube. If angle of contact of mercury is `140^@` and density `13.6 xx 10^3 kg m^(-3)` Surface tension of mercury is `540 xx 10^(-3)Nm^(-1)` |
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Answer» Correct Answer - `9.406 xx 10^(-4)m.` `h = 13.2 mm = - 13.2 xx 10^(-3) m , theta = 140^@` and `cos 140^@ = cos (180^@ - 40^@)` `=- cos 40^@ = 0.7660` `h =2 S cos theta//r rho g` ` or 2 r = 4S cos theta //h rho g` |
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| 316. |
What will be the effect on the angle of contact of a liquid if the temperature increases? |
| Answer» With the increase in temperature , the surface tension of liquid decreases. Due to it, the liquid surface on the solid surface becomes more flat, consequently the angle of contact of a liquid increases with the increase in temperature. | |
| 317. |
What angle will the free surface of a liquid subtend with the horizontal when the liquid in a container is moving horizontally with constat ac celeration a as shown in fig. |
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Answer» Consider a horizontal liquid cylinder in the liquid, of length L area of cross-section A. Let `h_(1)`and `h_(2)` be the heights of liquid column at the ends of cylinder fro the surface of liquid as shown in fig. for the translatory motion of cylinder with ac celeratoin a. `F_(1)-F_(2) = ma` or, `P_(1)A-P_(2)A = AL rho a` or `(h_(1)rho g - h_(2)rho g)A =AL rho a` `[ :. P = h rho g]` or `(h_(1)-h_(2))g = La` or `(h_(1)-h_(2))/(L) = a/g` From fig, ` tan theta = (h_(1)-h_(2))/(L) = a/g` so, `theta = tan^(-1) a//g`. |
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| 318. |
Why is mercury perferred as a barometric substance over water? |
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Answer» The pressure P exerted by column of a liquid of height h, density of liquid `rho is P = h rho g` or `h prop 1.(rho)` (for a constant P). Since mercury is a most dense liquid available. Therefore by using it the barometric arrangement will be of very comvenient size. To understand it, let the atmospheric pressure be `1.01 xx 10^(5)Pa`. Then teh height of water barometer is `h_(w) =(P)/(rho_(w)g) = (1.01 xx10^(5))/((13.6 xx 10^(3))xx 9.8) = 0.76 m` This indicates that the tube required for water barometer should be long. which is very inconvenient, whereas the tube used in mercury barometer is small, which is convenient to use. |
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| 319. |
Compare the densities of water at the surface and bottom of a lake 100 metre deep , given that the compressibility is `10^(-3)//22` per atmosphere and 1 atmosphere = `1.015 xx 10^(5)Pa`. |
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Answer» Compressibility = `1/B=(10^(-3))/(22) = (1)/922 xx 10^(3) `per atm `B = 22 xx 10^(3) atm = 22 xx 10^(3) xx 1.015 xx 10^(5) Pa` Let V be the volume of 1 kg water at surface and `(V-Delta V)` be its volume at the bottom of lake 100m deep. Increase in pressure `p = h rho g` `=100 xx 10^(3) xx 9.8 Nm^(-2)` `(Delta V)/(V) = p/B = (100 xx 10^(3) xx 9.8 )/(22 xx 10^(3) xx (1.015 xx 10^(5)))` `("Density of water at the surface")/("density of water at the bottom") = (1//V)/(11//(V-Delta V))` `=(V-Delta V)/(V) = 1-(Delta V)/(V) = 1 -(9.8)/(22xx1.015 xx 10^(3))` `=0.99956`. |
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| 320. |
Solar pond is a device for collecting solar heat. The pond is about one metre deep, filled with saturated salt solution and protected from air current and other disturbances. When exposed to sun, the temperature at the bottom can go as high as `80^(@)C` or more. why is this possible? |
| Answer» It is due to the fact that the thermal conductivity of salt is very high. | |
| 321. |
Water rises to a height of 10 cm. in a certain capillary tube. If in the same tube, level of Hg is depressed by 3.42 cm., compare the surface tension of water and mercury. Sp. Gr. Of Hg is 13.6 the angle of contact for water is zero and that for Hg is `135^@.` |
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Answer» Correct Answer - 0.152 Let `S_1, S_2` be the surface tension of water and mercury respectively. Let h be the height of water raised due to S.T. and `h_2` the depression of mercury. Then `h_1 = 10 cm , h_2 = - 3.42 cm.` Angle of contant for water and capillary tube, `theta_1 = 0^@` and for mercury and capillary tube, `theta_2 = 135^@.` `h_1 = (2S_1 cos theta_1)/(r rho_1g) and h_2 = (2S_2 cos theta_2)/(r rho_2g)` `:.(h_1)/(h_2) = (S_1 cos theta_1rho_2)/(S_2 cos theta_2 rho_1) or (S_1)/(S_2) = (h_1 rho_1 cos theta_2)/(h_2 rho cos theta_1)` `or (S_1)/(S_2) = (10 xx 1xxcos 135^@)/((-3.42)xx 13.6 xx cos 0^@)` `=(10xx(-0.7071))/((-3.42)xx 13.6) = 0.152` |
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| 322. |
A liquid of density `rho` and surface tension S rises to a height h in a capillary tube of diameter D. what is the weight of the liquid in the capillary tube? Angle of contact is `0^(@)`. |
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Answer» Weight of liquid in the capillary tube is `W=mg = pi r^(2)h rho g` We know that, `h=(2 S cos theta)/(r rho g) = (2S)/(r rho g)` `:. W = pi r^(2) rho g xx (2S)/(r rho g) = pi xx 2r xx S` `= pi D S`. |
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| 323. |
A capillary tube when immersed vertically in a liquid records a rise of 3cm.if the tube is immersed in the liquid at an angle of `60^(@)` with the vertical, then find the length of the liqiud column along the tube. |
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Answer» Vertical height of the liquid column `h (=3cm)` in inclined capillary tube will be the same as that in the vertical capillary tube when immersed in a liquid. If l is the length of liquid column along the tube in the inclined tube, then ` cos theta = h/l or l= (h)/(cos theta) = 3/ (cos theta 60^(@)) = 6 cm `. |
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| 324. |
When a capillary tube of radius r is immersed in a liquid of density `rho,` the liquid rises to a height h in it. If m is the mass of the liquid in the capillary tube, find the potential energy of this mass of the liquid in the tube. |
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Answer» Correct Answer - `mgh//2` The centre of gravity of mass m of the liquid in the capillary tube will be at height h/2 from the surface of liquid where h is the height through which liquid rises in the tube. Therefore, potential energy of liquid in the tube =mgh/2. |
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| 325. |
Two capillary tubes of equal length and inner radii 2r and 4r respectively are added in series and a liquid flows through it. If the pressure difference between the ends of the whole system is 8.5 cm of mercury, find the pressure difference between the ends of the first capillary tube. |
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Answer» Let `P , P_(1), P_(2)` be the pressure at the beginning of a tube, at the joint of two tubes and at the end of second tube. Since the two tubes are in series, so `(pi(P-P_(1))(2r)^(4))/(8eta l) = (pi(P_(1)-P_(2))(4r)^(4))/(8eta l)` or `P-P_(1) = (P_(1)-P_(2)) 16` or `P=17 P_(1) = 16 P_(2)` ...(i) Given, `P-P_(2) = 8.5` ...(ii) Fron (i) and (ii) , we have `P_(2) +8.5 = 17 P_(1) - 16P_(2)` or `P_(1)-P_(2) = 8.5//17 = 0.5 cm" of "Hg`. `:.` pressure difference between the ends of the first tube `=P-P_(1)=(P-P_(2)) - (P_(1)-P_(2))` `=8.5 - 0.5 = 8.0" cm of "Hg`. |
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| 326. |
Water rises in a capillary tube to a height 2.0 cm. In an another capillary tube whose radius is one third of it, how much the water will rise ? If the first capillary tube is inclined at an angle of `60^@` with the vertical then what will be the position of water in the tube. |
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Answer» Correct Answer - 6.0 cm ; 4.0 cm `h = 2 S cos theta//r rho g` or `h r = 2 S cos theta//rho g =` a constant `:. H_1 r_2 = hr or h_1 = hr//r_1 = hr//(r//3)` `= 3h = 3xx 2.0 = 6.0 cm` Let length of water in the capillary tube be `h_2` in the inclined position, the vertical height h ( = 2.0 cm) of the liquid will remain the same. Then `h_2 = h//cos 60^@ = 2(1//2) = 4.0 cm` |
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| 327. |
If two capillary tubes of radii `r_1 and r_2` and having length `l_1 and l_2` respectively are connected in series across a heaed of pressure p, find the rate of flow of the liqid through the tubes, if `eta` is the coefficient of viscosity of the liquid. |
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Answer» Correct Answer - `(pi P)/(8 eta(l_1//r_1^4 + l_2//r_2^4))` Let `P_(1)` and `P_(2)` be the pressure difference across the first and second capillary tube respectively. Let `R_(1), R_(2)` be the second capiliary tube respectively. Then `R_(1)=(8 etal_(1))/(pi r_(1)^(4))` and `R_(2)=(8 etal_(2))/(pir_(2)^(4))` As the two capillary tubes are connected in series total liquid resistance, `R=R_(1)+R_(2)=(8 eta l_(2))/(pi r_(1)^(4))=(8 eta)/(pi)[(l_(1))/(r_(1)^(4))+(l_(2))/(r_(2)^(4))]` Rate of flow loquid through the tubes is `V=(P)/(R)=(P)/(8 eta)/(pi)[(l_(1))/(r_(1)^(4))+(l_(2))/(r_(2)^(4))]=(pi P)/(8 eta[(l_(1))/(r_(1)^(4))+(i_(2))/(r_(2)^(4))]` |
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| 328. |
Due to change in main voltage, the temperature of an electric bulb rises from 3000K to 4000K. What is the percentage rise in electric power consumed? |
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Answer» When temperature , `T_(1) = 3000K` then `E_(1) = sigma T_(1)^(4) = xx sigma (3000)^(4)`..(i) When temperature, `T_(2) = 4000K`, then `E_(2) = sigma T_(2)^(4) = xx sigma (4000)^(4)`..(ii) % rise in electric power is `=(E_(2)-E_(1))/(E_1) xx100 = ((E_2)/(E_1)-1)xx100` `=(((4000)^(4))/((3000)^(4))-1)xx100 = ((256)/(81)-1)xx100 = 216%`. |
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| 329. |
Two rods A and B have lengths `l_1 and l_2.` Each rod has its ends at temperature `T_1 and T_2` Radii fo cross section of the two rods are same. The condition for equal flow of heat through those two rods is `(K_1, K_2` are thermal conductivities of two rods)A. `K_1 I_1 =K_2 I_2`B. `K_1 I_2 =K_2 I_1`C. `K_1 = K_2`D. `I_1= I_2` |
| Answer» Correct Answer - B | |
| 330. |
A body of mass 6 kg is floating in a liquid with `2//3` of its volume inside the liquid. Find (i) buoyant force acting on the body and (ii) ratio between the density of the body and density of liquid. Take `g=10m//s^(2)`. |
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Answer» when a body floats, the buoyant force on the body is equal and opposite to the weight of body. So, Buoyant force = weight of body `= 6 kg xx 10m//s^(2)=60N` Let V be the volume of the body. Buoyant force = wt. of liquid displaced `((2)/(3)V) rho_(l)g` Weight of body `= Vrho_(b) g` when body floating, then `V rho_(b) g = 2/3 V rho_(l) g or (rho_b)/(rho_l) = 2/3`. |
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| 331. |
Why burns from steam are more serious than those from boling water? |
| Answer» The steam at `100^(@)C` carries `22.6 xx 10^(5) J kg^(-1)` more heat than water at boiling point `100^(@)C` . That is why the burns from the steam are generally more serious than those from boling water. | |
| 332. |
A sphere of alumininum of mass 0.047 kg placed for sufficient time in a vessel containing boling water, so that the sphere is at `100^(@)C`. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at `20^(@) C` . The temperature of water rises and attains a steady state at `23^(@)C` . calculate the specific heat capacity of aluminum. Specific heat capacity of copper = `0.386 xx 10^(3) J kg^(-1) K^(-1)`. Specific heat capacity of water = `4.18 xx 10^(-3) J kg^(-1) K^(-1)` |
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Answer» Here, `m_(Al) = 0.047 kg` , `T_(1) = 100^(@)C , m_(cu) = 0.14 kg` , `m_(w) = 0.25 kg , T_(0) = 20^(@)C` `T_(2) = 23^(@) C , s_(Cu) = 0.386 xx 10^(3) J kg^(-1) K^(-1)` Heat lost by aluminium, `Q_(1) = m_(Al)s_(Cu) (T_(1)-T_(2))` `=0.047 xx s_(Al) xx (100 -23)` `=0.047 xx s_(Al) xx 77 J` Heat taken by copper calorimeter and water is `Q_(2) = m_(Cu) s_(Cu) (T_(2)-T_(0))+m_(w)s_(w)(T_(2)-T_(0))` `= 0.14 xx (0.386 xx 10^(3)) xx (23 -20)` `+0.25 xx (4.18 xx 10^(3)) (23-20)` `= 162.12 + 3135 = 3197.12 J` In the steady state,haet lost= heat gained `:. 0.04 xx s_(Al) xx 77 = 3297.12` or, `s_(Al) = (3297.12)/(0.047 xx 77) = 911 J kg^(-1)K^(-1)`. |
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| 333. |
The pressure of a medium is charged from `1.01 xx 10. Pa` to `1.165 xx 10. Pa` and change in volume is 10 % keeping temperature constant. The Bulk modulus of the medium isA. `204.8 xx 10^5 Pa`B. `102.4 xx 10^5 Pa`C. `51.2xx10^5 Pa`D. `1.55 xx 10^5 Pa` |
| Answer» Correct Answer - D | |
| 334. |
At what speed will the velocity head of a stream of water be equal to 20 cm of mercury . Taking `(g=10ms^(-2))`. |
| Answer» Refer to solve Example `20`. | |
| 335. |
Two metal cubes A and B of same size are arranged as shown in Figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are `300W//m ^@C and 200 W//m^@C,` respectively. After steady state is reached the temperature t of the interface will be ...... |
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Answer» Correct Answer - `[60^(@)C]` At steady state Rate of flow of heat through cube A = rate of flow of heat through cube B `:. (K_(1)A(100-T))/(x)=(K_(2)A(T-0))/(x)` or `(300A (100-T))/(x)=(200A(T-0))/(x)` or `300(100-T)=200T` or `300-3T=2T` or `T=60^(@)C` |
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| 336. |
in river, the deep water runs…………. And less deep water runs………… |
| Answer» Correct Answer - slow, fast | |
| 337. |
A thin rod, length `L_(0)` at `0^(@)C` and coefficient of linear expansion `alpha` has its two ends mintained at temperatures `theta_(1)` and `theta_(2)` respectively Find its new length . |
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Answer» Consider the temperature of the rod varies linearly from one end to other end. Let `theta` be the temperature of the rod at the mid point of rod. At steady state `(dQ)/(dt) = (KA(theta_1 -theta))/(L_0//2) = (KA(theta- theta_2))/(L_0//2) or theta_1 - theta = theta - theta_2 or theta = (theta_1 + theta_2)/(2)` Using, `L = L_0 (1 +alpha theta),` We have `L = L_0[1 + alpha((theta_1 + theta_2)/(2))]` |
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| 338. |
When 0.15 kg of ice at `0^(@)C` is mixed with 0.30 kg of water at `50^(@)C` in a container, the resulting temperature is `6.7^(@)C`. Calculate the heat of fusion of ice. `(s_(water) = 4186 J kg^(-1)K^(-1))` |
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Answer» Heat lost by water `=m_(w)s_(w) (T_(1)-T_(2)) = 0.30 xx 4186 xx (50-6.7)` `= 54376.14 J` Heat taken by ice =`m_(i)L + m_(i)s_(w)(T_(2)-T_(0))` `=0.15 xx L + 0.15 xx 4186 xx (6.7-0)` `=0.15 L + 4206.93 J` Heat lost = heat gained `:. 54376.14 = 0.15 L + 4206.93` or `L = 3.34 xx 10^(5) J kg^(-1)`. |
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| 339. |
A cloudy night is ………… then a ………… night. |
| Answer» Correct Answer - warmer , clear sky | |
| 340. |
Two equal drops of water falling through air with a steady velocity `5 cm//s`. If the drops combire to from a single drop, what will be new terminal velocity? |
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Answer» Here, `v=5 cms^(-1)` .Let r be the radius of each drop. Terminal velocity of each drop is `v=(2pi r^(2)(rho-sigma)g)/(9 eta)=5` ……(i) Let R be the radius of by drop formed when two small drop coalese together. Then `(4)/(3) piR^(3) =2xx(4)/(3) pi r^(3)` or `R=2^(1//3) r` Terminal velocity of big drop is `v=(2R^(2)(rho- sigma)g)/(9 eta)` .....(ii) Dividing (ii) by (i), we get `(V)/(5) =(R^(2))/(r^(2))` or `V=5xx(R^(2))/(r^(2))` `V=5xx((2^(1//3)r)^(2))/(r^(2))=5xx2^(2//3)=5xx1.5874` `=7.937 cms^(-1)` |
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| 341. |
Woolen cloths are worn in winter but not in summer. Why? |
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Answer» The fiber of woolen clothes enclose a large number of air cavities in them. Since both air and wool being bad conductor of heat prevent the loss of heat prevent the loss of heat by conduction from our body to surrounding. So we feel warm in winter in woolen cloths. In summer , the temperature of surrounding is higher than the temperature of our body. if we wear the woolen clothes. then the heat from surrounding reaches to body through radiation but not through conduction. But heat can not flow from our body to surrounding due to conduction. As a result of it, our body temperature becomes more than normal and we feel discomfort. |
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| 342. |
The heat is supplied to a given mass of water. Draw the variation of volume of water with its temperature. |
| Answer» The variation of given volume of one kg of water (V) with temperatrue `(t^(@)C)` . | |
| 343. |
Density of solid decrease/increases with rise in temperature. Explain. |
| Answer» For most of solids, the volume increases with the rise in temperature. As density = mass//volume, therefore, for a given mass, density of a solid decreases with rise in temperature. | |
| 344. |
if a ball of steel (density `rho=7.8 g//cm^(3)`) attains a terminal velocity of `10 cm//s` when falling in a tank of water (coefficient of viscosity, `eta_(water) =8.5xx10^(-4)`Ps s), then its terminal velocity in glycerine `(rho =1.2 g//cm^(2), eta =13.2 Pa s)` would be nearlyA. `6.45xx10^(-4)cm//s`B. `1.5xx10^(-5)cm//s`C. `1.6xx10^(-5)cm//s`D. `6.25xx10^(-4)cm//s` |
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Answer» Correct Answer - D Here, `rho_(s)=7.8 g//c^(3), v_(1) =10 cm//s`, `eta_(water) =8.5xx10^(-4)Pa S` `rho_(w)=1, rho_(g) =1.2 g//cm^(3), eta_(g) =13.2 Pa S, v_(2)=?` As terminal velocity `v=(2r^(2)(rho-rho_(0))g)/(9eta)` `:. v prop((rho-rho_(0)))/(eta)` `(v_(2))/(v_(1))=((rho-rho_(g)))/(eta_(g))xx(eta_(water))/((rho-rho_(w)))` `=((7.8-1.2))/(13.2)xx(8.5xx10^(-4))/((7.8-1)) = (6.6xx8.5xx10^(-4))/(13.2xx6.8)` `(v_(2))/(v_(1)) =0.625xx10^(4)` or `v_(2) =0.625xx10^(-4) xxv_(1)` `v_(2)=0.625xx10^(-4)xx10= 6.25xx10^(-4) cm//s` |
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| 345. |
Work done in increasing the size of a soap bubble from a radius of 3cm to 5cm is nearly (Surface tension of soap solution `=0.03Nm^-1`)A. `0.4 pi mj`B. `4 pi mj`C. `0.2 pi mj`D. `2 pi mj` |
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Answer» Correct Answer - A Here `S=0.03 N//m, r_(1)=3cm=3xx10^(-2)m`, `r_(2)=5 cm=5xx10^(-2)m` Work done (W) in increasing the size of soap bubble = surface tension x increase in area of both the free surface of the bubble `W =Sxx(4pi R_(2)^(2)-4pi R_(1)^(2)) =8pi S[R_(2)^(2)-R_(1)^(2)]` `=8pi(0.03) [(5xx10^(-2))-(3xx10^(-2))^(2)]` `=3.84xxpixx10^(-4)j=4pixx10^(-4)j= =.4pi mj` |
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| 346. |
A lead shot of 1 mm diameter falls through a long colummn of glycerine. The variation of the velocity v with distance covered (s) is represented byA. B. C. D. |
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Answer» Correct Answer - A In the beginning, due to gravity pull, the lead shot will be ac celerated and hence will move with increasing velocity for some time. When the viscous force balance the gravity pull, then the shot will move with constant velocity by shot. Thus option (a) is correct. |
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| 347. |
Why does a metal bar appear hotter then a wooden bar at the same temperature ? Equivalently it also appears cooler then wooden bar if they are both colder then room temperature. |
| Answer» Metal bar is good conductor of heat then wooden bar. When we touch by hand a hot metal bar and wooden bar (at the same temperaure), then more heat flows from metal bar to out body then that due to wooden bar. Due or it the metal bar apperar hotter then wooden bar. Due to it, the metal bar appears colder then wooden bar. | |
| 348. |
A rigid bar of mass `M` is supported symmetrically by three wires each of length `l`. Those at each end are of copper and the middle one is of ion. The ratio of their diameters, if each is to have the same tension, is equal toA. `(Y_(copper))/(Y_(iron))`B. `sqrt((Y_(iron))/(Y_(copper)))`C. `(Y_(iron)^(2))/(Y_(copper)^(2))`D. `(Y_(iron))/(Y_(copper))` |
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Answer» Correct Answer - B `Y=(F//pi(D//2)^(2))/(Delta l//l) = (4 F l)/(pi D^(2) Delta l)` `:. D=sqrt((4F l)/(pi Delta l Y))`. or `D prop sqrt((1)/(Y)` Hence, `(D_(copper))/(D_(iron)) = sqrt((Y_(iron))/(Y_(copper))` |
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| 349. |
A large block of ice `5 m` thick has a vertical hole drilled through it and is floating in the middle of a lake. What is the minimum length of a rope required to scoop up a bucket full of water through the hole? Relative density of ice `=0.9`. |
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Answer» Let `rho` and `rho_(1)` be the density of ice and see water respectively. so, `rho//rho_(1)=0.9` Let H be the thickness of ice block, of crosso section area A. The weight of ice block = `A H rho g` Let x be the thickness of iceblock above the surface of water `:.` Thickness of ice block inside the water `=(H-x)` By the principle of floatation, `AH rhog=(H-x) A rho_(1) g` or `x=H(1-rho//rho_(1)) = 5(1-0.9)=0.5 m`. |
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| 350. |
A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` mm^(2)`, respectively . At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Given, `Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-2)` |
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Answer» For steel wire `A, l_(1) = l, A_(1) = 1 mm^(2) , Y_(1) = 2 xx 10^(11)Nm^(-2)` For aluminium wire B, `l_(2) = l , A_(2) = 2 mm^(2) , Y_(2) = 7 xx 10^(10) Nm^(-2)` (a) Let mass m be suspeneded from the rod at distance x from the end where wire A is connected. let `F_(1)` and `F_(2)` be the tensions in two wires and there is equal stress in two wires, then `(F_1)/(A_1) = (F_2)/(A_2) or (F_1)/(F_2) = (A_1)/(A_2) = 1/2 ` ..(i) Taking moment of forces about point of suspension of mass from the rod, we have `F_(1) x = F_(2) (1.05 -x) or (1.05 = x)/(x) = (F_1)/(F_2) = 1/2` or ` 2.10 -2x = x or x=0.70 m = 70 cm`. (b) Let mass m be suspeneded from the rod at distance x from the end where wire A is connected. let `F_(1)` and `F_(2)` be the tensions in two wires and there is equal stress in two wires, ie. , `(F_1)/(A_(1)Y_(1)) = (F_2)/(A_(2)Y_(2))` or `(F_1)/(F_2) = (A_(1)Y_(2))/(A_(2)Y_(2)) = 1/2 xx (2 xx 10^(11))/(7 xx 10^(10)) = 10/7` As the rod is stationary, os `F_(1) x = F_(2) (1.05-x) or (1.05 = x)/(x) = (F_1)/(F_2) =10/7` or, `10x = 7.35 - 7x or x=0.4324m = 43.2cm`. |
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