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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.
| 251. |
A plano convex lens is made of glass of refractive index 1.5. The radius of curvature of its convex surface is R. Its focal length isA. `R/2`B. RC. 2RD. 1.5R |
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Answer» Correct Answer - C `1/f(1.5-1)[1/R-1/(oo)]` `f=R/0.5` f=2R |
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| 252. |
Calculate the focal length of a convex lens of crown glass of dispersive power 0.012 and concave lens of dispersive power 0.020 that from an achromatic coverging combination of focal length 0.3 m when placed in contact. |
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Answer» Correct Answer - 12 cm for convex lens, 20 cm for concave lens |
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| 253. |
Heat given to a body which raises its temperature by `1^@C` isA. water equivalentB. thermal capacityC. specific heatD. temperature gradient |
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Answer» Correct Answer - B Amount of heat required by a body of any mass of undergo a unity change in temperature is known as heat capacity or thermal capacity of the substance. |
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| 254. |
An aluminium sphere of `20cm` diameter is heated from `0^(@)C` to `100^(@)C`. Its volume changes by (given that the coefficient of linear expanison for aluminium `(alpha_(Al) = 23 xx 10^(-6//0)C)`A. `28.9` ccB. `2.89` ccC. `9.28` ccD. `49.8` cc |
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Answer» Correct Answer - A `DeltaV = V_(0)(3alpha)Deltatheta` =`(4)/(3)(3.14)(10)^(3)[3 xx 23 xx 10^(-6)][100-0]` =`28.9` cc |
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| 255. |
Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in figure (a) & (b) . In arrangement (a) light converges at point D , which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m-1) from C. Find out the refractive index of the material. |
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Answer» Correct Answer - A For case (a) `(mu)/(v_(1)) - (1)/(oo) = (mu-1)/(R), (1)/(v_(2))- (mu)/(v_(1) - R) = (1-mu)/(oo) "and" v_(2) = (2R)/(m) implies m=2(mu-1)mu` For case(b) `(1)/(v_(2))- (mu)/(oo) = (1-mu)/(-R) "and" v_(2) = (R)/(m-1) implies mu=m` Therefore `mu= 2 (mu-1) mu implies mu-1 = (1)/(2) implies mu = 1.5` |
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| 256. |
A quarter cylinder of radius R and refractive index 1.5 is placed on a table.A point object P is kept at a distance of mR from it. Find the value of m for whicha ray from P will emerge parallel to the table as shown in the figure. |
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Answer» Correct Answer - `4//3` Applying `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R)` First on plane surface `(1.5)/(Al_(1)) - (1)/(-mR) = (1.5-1)/(oo) = 0` `therefore Al_(1) = -(1.5"mR")` Then on curved surface `(1)/(oo) - (1.5)/(-(1.5 mR + R)) = (1 - 1.5)/(-R)` [`v = oo` because final image is at infinity] `implies (1.5)/((1.5 m +1) R) = (0.5)/(R)` `implies 3 = 1.5 m + 1 implies (3)/(2) m = 2 implies m = (4)/(3)` |
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| 257. |
A gram molecule of a gas at `127^(@)C` expands isothermally until its volume is doubled. Find the amount of work done and heat absorbed. |
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Answer» n=`2` , `T=127+273 = 400K, (V_(2))/(V_(1)) = 2 ` From formula `" "` `W = 2.3026 nRT "log"_(10) (V_(2))/(V_(1)) = 2.3026xx2xx8.3xx400xx"log"_(10)2` ` = 2.3026 xx2xx8.3xx400xx0.3010 approx 4.6 xx 10^(3)J` |
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| 258. |
The internal energy of a system remains constant when it undergoes (i) a cyclic process (ii) an isothermal process (iii) an adiabatic process (iv) any process in which the heat given out by the system equal to the work done on the systemA. a cyclic processB. an isothermal processC. an adiabatic processD. any process in which the heat given out by the system is equal to the work done on the system |
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Answer» Correct Answer - A::B::D `Q= DeltaU + W` and `DeltaU = nC_(v)DeltaT` `DeltaU` can be zero if `DeltaT` is zero or `Q - W` is zero |
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| 259. |
One mole of an ideal monoatomic gas undergoes a cyclic process as shown in figure . Temperature at point `1` =`300K` and process `2-3` is isothermal . The efficiency of cycle isA. `((9ln3+4)/(9ln3+ 12))`B. `((9ln3-4)/(9ln3+ 12))`C. `((9ln3+4)/(9ln3+ 16))`D. `((9ln3+12)/(9ln3+ 16))` |
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Answer» Correct Answer - C `eta = (W)/(DeltaQ_(123)) = ((9"ln"3-4)/(9"ln"3+16))` |
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| 260. |
One mole of an ideal monoatomic gas undergoes a cyclic process as shown in figure . Temperature at point `1` =`300K` and process `2-3` is isothermal . Heat capacity of process `1rarr 2` isA. `(R)(2)`B. `(3R)/(2)`C. `(5R)/(2)`D. `2R` |
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Answer» Correct Answer - D `C = (DeltaQ)/("n"DeltaT) = (n((3R)/(2)DeltaT) +DeltaW)/(nDeltaT) = (3R)/(2) + (4P_(0)V_(0))/("n"DeltaT) = (3R)/(2) + (P_(0)V_(0))/(n(2T_(0))) = (3R)/(2) + (R)/(2) = 2R` `DeltaQ = n(2R)(DeltaT_(12)) + (W_(23)) = n(2R)(8T_(0)) + nR(9T_(0))ln(3) = nRT_(0) (16 +9ln3)` |
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| 261. |
One mole of an ideal monoatomic gas undergoes a cyclic process as shown in figure . Temperature at point `1` =`300K` and process `2-3` is isothermal . . Net work done by gas in complete cycle isA. `(9ln3 +12)P_(0)V_(0)`B. `(9ln3+4)P_(0)V_(0)`C. `(9ln3-4)P_(0)V_(0)`D. `(9ln3 - 8)P_(0)V_(0)` |
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Answer» Correct Answer - C `W = W_(12) + W_(23) + W_(31) = (1)/(2)(4P_(0)) (2V_(0)) + "n"RT"ln"(3) - P_(0) (8V_(0)) = -4P_(0)V_(0) + 9 P_(0)V_(0)ln3 = (9ln3-4)P_(0)V_(0)` |
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| 262. |
A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screenA. Half the image will disappearB. Complete image will be formed of same intensityC. Half image will be formed of same intensityD. Complete image will be formed of decreased intensity. |
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Answer» Correct Answer - D Because to form the complete image only two rays are to be passed through the lens and moreover, since the total amount of light released by the object is not passing through the lens, therefore image is faint ( intensity is decreased ) |
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| 263. |
A ray of light is incident normally on a plane mirrorr. The angle of reflection will beA. `0^(@)`B. `90^(@)`C. Will not be reflectedD. None of these |
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Answer» Correct Answer - A ` lt i= lt r = 0^(@)` |
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| 264. |
STATEMENT -1: There exists two angles of incidence for the same deviation (except minimum deviation) by a prism kept is air. STATEMENT-2: This is due to principal of reversibility of lightA. Statement-1 is true, Statement-2: is true, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is true, Statement-2: is true, Statement-2 is NOT a correct explanation for Statement-1.C. Statement-1 is true but statement-2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A In case of minimum deviation i=e, there exists only one angle of incidence. |
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| 265. |
All of the following statements are correct exceptA. The magnification produced by a convex mirrorr is always less than oneB. A virtual, erect, same-sized image can be obtained using a plane mirrorrC. A virtual, erect, magnified image can be formed using a concave mirrorrD. A real, inverted, same-sized image can be formed using a convex mirrorr |
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Answer» Correct Answer - D Convex mirrorr always forms, virtual, erect and smaller image |
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| 266. |
Assertion `:` For the sensitivity of a camera, its aperture should be reduced Reason `:` Smaller the aperture, image focussing is also sharp.A. If both the assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C Very large apertures gives blurred images because of aberrations. By reducing the aperture the clear image is obtained and thus the sensitivity of camera increases. Also the focussing of object at different distance is achieved by slightly altering the separation of the lens from the film . |
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| 267. |
Assertion `:` The size of the mirrorr affect the nature of the image. Reason `:` Small mirrorrs always forms a virtual image.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D The size of the mirrorr does not affect the nature of the image except that a bigger mirrorr forms a brighter image. |
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| 268. |
`1500` ml of a gas at a room temperature of `23^(@)C` is inhaled by a person whose body temperature is `37^(@)C`, if the pressure and mass stay constant , what will be the volume of the gas is the lungs of the person? |
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Answer» `T_(1) =273 +37=310 K, T_(2) = 273+23 = 296 K`. Pressure and amount of the gas are kept constant, So `(V_(1))/(T_(1))= (V_(2))/(T_(2)) therefore V_(2) = V_(1)xx (T_(2))/(T_(1)) = 1500 xx (293)/(310) = 1417.74` ml |
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| 269. |
A sample of `O_(2)` is at a pressure of `1` atm when the volume is `100` ml and its temperature is `27^(@)C`. What will be the temperature of the gas if the pressure becomes `2` atm and volume remain`100`ml. |
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Answer» `T_(1) =273+27 = 300K` For constant volume `P_(1)/T_(1)= P_(2)/T_(2) implies T_(2) = T_(t)xx P_(2)/P_(1) = 300xx2/1 = 600 K = 600-273= 327^(@)C` |
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| 270. |
The masspercentage in composition of dry air at sea level contains contains approximately `75.5%` of `N_(2)`. If the total atmosphere pressure is `1` atm then what will be the partial pressure of nitrogen? [Given : average molecular mass of air = `29`] |
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Answer» The mole fraction of nitrogen `mu_(1) = ((M_("nitrogen"))/("Molecular weight")) = 0.755xx(29)/(28) = 0.78` The partial pressure of nitrogen `P_(1) = mu_(1)(RT)/(V) = (mu_(1))/(mu) (muRT)/(V) =(mu_(1)/ mu)P = (0.78)xx1 = 0.78`atm |
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| 271. |
Statement I: A ray is incident from outside on a glass sphere surrounded by air as shown in Figure. This ray may suffer total internal reflection at the second interface. Statement II: For a ray going from a denser to rarer medium, the ray may suffer total internal reflection.A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B When a ray of light enters a spherical glass sphere, it is first refracted at first interface and then it strikes the inner surface of sphere ( second interface ) and get totally internally relected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle `(42^(@)`, in this case ). |
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| 272. |
Wavelength of light used in an optical instrument are `lambda_(1)=400 Å ` and `lambda_(2)=5000 Å`, then ratio of their respective resolving power (corresponding to `lambda_(1)` and `lambda_(2))` isA. `16:25`B. `9:1`C. `4:5`D. `5:4` |
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Answer» Correct Answer - D `R.P.=(1)/(lambda)implies((R.P.)_(1))/((R.P.)_(2))=(l_(2))/(l_(1))=(5)/(4)` |
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| 273. |
A linear object AB is placed along the axis of a concave mirror. This object is moving towards the mirror with speed U. The speed of the image of the point A is 4U and the speed of the image of B is alSO 4U btu in opposite direction. If the center of the line AB is at a distance L from the mirror then find out the length of the object. A. `3L//2`B. `5L//3`C. LD. None of these |
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Answer» Correct Answer - c. For concave mirror, if x and y are object distance and image and distance, recpectivley, we have `-(1)/(x)-(1)/(y)=-(1)/(|f|)` `rArr(1)/(x)+(1)/(y)=(1)/(|f|)` `rArr-(1)/(x)(dx)/(dt)-(1)/(y)(dy)/(dt)=0` `rArr |(V_(x))/(V_(y))|=(x^(2))/(y^(2))` For`rArr |(V_(x))/(V_(y))|=(1)/(4),(x)/(y)=+-2` For `(x)/(y)=2` we get `x=(3|f|)/(2)` [for point A] and For`(x)/(y)=-2`, we get `x=(|f|)/(2)` [for point B] As the middle happens to be focus of the mirror, we get `|f|=L` |
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| 274. |
A mango tree is at the bank of a river and one of the branch of tree extends over the river. A tortoise lives in the river. A mango falls just ono the tortoise. The acceleration of the mango falling from tree as it appears to the tortoise is (refractive index of water is `4//3` and the tortoise is stationary)A. gB. `3g//4`C. `4g//3`D. none of these |
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Answer» Correct Answer - c. `(x)/(1)=(x_(rel))/(mu)rArr X_(rel)=mux` `(d^(2)x_(rel))/(dt^(2))=mu(d^(2))/(dt^(2))rArra_(rel)=mug` |
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| 275. |
The exposure time of a camera lens at the `(f)/(2.8)` setting is `(1)/(200) ` second. The correct time of exposure at `(f)/(5.6)` isA. `0.4 sec`B. `0.02 sec`C. `0.002 sec`D. `0.04 sec` |
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Answer» Correct Answer - B Time of exposure `prop (f.n umber)^(2)` `implies(t_(2))/(f_(1))=((5.6)/(2.8))^(2)=4` `t_(2)=4t_(1)=4xx(1)/(200)=(1)/(50)sec=0.02 sec.` |
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| 276. |
A thin lens produces an upright image of the same size as the object. Then from the optical centre of the lens, the distance of the object is .A. ZeroB. `4f`C. `2f`D. `(f)/(2)` |
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Answer» Correct Answer - A `m=(f)/(f+u)` `1=(f)/(f+u)`or `f+u=f` or `u=0` or `u=0` |
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| 277. |
A convex lens of focal length `10cm` and concave lens of focal length `20 cm` are kept `5 cm` apart. The focal length of the equivalent lens isA. `(120)/(3)cm`B. `18cm`C. `30 cm`D. `(40)/(3)` |
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Answer» Correct Answer - D `(1)/(F)=(1)/(10)+(1)/(-20)-(5)/(10(-20))` `(1)/(F)=(1)/(10)-(1)/(20)+(5)/(200)` `(1)/(F)=(20-10+5)/(200)`or `F=(200)/(15)cm=(40)/(3)cm` |
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| 278. |
A microscope has an objective of focal length `1.5 cm` and eye piece of focal length `2.5 cm`. If the distance between objective and eyepiece is `25 cm`. What is the approximate value of magnification produced for relaxed eye ?A. 75B. 110C. 140D. 25 |
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Answer» Correct Answer - C Length of the tube is `L=v_(o)+f_(e)` `v_(o)=L-f_(e)` Now applying `(1)/(v_(o))-(1)/(u_(o))=(1)/(f_(o))` we have `(1)/(22.5)-(1)/(u_(o))=(1)/(1.5)` `:. |u_(o)| ~~1.6cm` `:. |M|= (v_(o))/(u_(o))xx(D)/(f_(e))` `=((22.5)/(1.6))((25)/(2.5)) ~~140~~140` |
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| 279. |
In a compound microscope the objective of `f_(o)` and eyepiece of `f_(e)` are placed at distance `L` such that `L` equalsA. `f_(o)+f_(e)`B. `f_(o)-f_(e)`C. Much greater than `f_(o)` or `f_(e)`D. Need not depend either value of focal lengths |
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Answer» Correct Answer - C `L=v_(o)+u_(e)` and `v_(o)gt gtf_(o),u_(e)~=f_(e)` |
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| 280. |
The focal lengths of the objective and eye-piece of a compound microscope are 1 cm and 5 cm, respectively. An object is placed 11 mm from the objective and the final image is 25 cm from the eye. Find : (a) magnification produced and (b) the separation of the lenses. |
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Answer» Correct Answer - `66,15.1 cm` |
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| 281. |
A compound microscope has an objective of focal length 2 cm and eye-piece of focal length 5 cm. The distance between the two lenses is 25 cm. If the final image is at a distance of 25 cm from the eye-piece, find the magnifying power of the microscope. What would be the magnifying power if the microscope were reversed ? |
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Answer» Correct Answer - 56.5 |
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| 282. |
When a system is taken from state a to state b, in figure along the path `a to c to b`, `60J` of heat flow into the system, and `30 J` of work is done : (i) How much heat flows into the system along the path `a to d to b` if the work is `10J`. (ii) When the system is returned from b to along the curved path, the work done by the system is `-20J`. Does the system absorb or liberate heat, and how much ? (iii) If , `U_(a) = 0` and `U_(d)= 22J`, find the heat absorbed in the process `a to d ` and `d to b`. |
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Answer» For the path a, c, b, `DeltaU=Q-W = 60-30= 30J` or `U_(b) - U_(a) = 30J` (i) Along the path a, d, b, ` Q=DeltaU + W = 30+10= 40J` (ii) Along the curved path b, a, `Q= (U_(a)-U_(b)) + W = (-30) + (-20) = -50J`, heat flows out the system (iii) `Q_(ad) = 32J, Q_(db) = 8J` |
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| 283. |
One end of a brass rod `2m` long and having `1cm` radius is maintained at `250^(@)C`. When a steady state is reached , the rate of heat flow acrss any cross-section is `0.5 cal s^(-1)`. What is the temperature of the other end `K=0.26"cal" s^(-1) cm^(-1). ^(@)C^(-1)`. |
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Answer» `Q/t = 0.5 cal s^(-1), r=1cm` ` therefore Area A =pir^(2)= 3.142xx1cm^(2)= 3.142 cm^(2)` L=length of rod =`2m = 200cm, T_(1) = 250^(@)C, T_(2)=?` We know `Q/t = (KA(T_(1)-T_(2)))/L ` or `(T_(1)-T_(2))= Q/txx(Deltax)/(kA)= (0.5xx200)/(0.26C^(-1)xx3.142) = 122.4^(@)C` `therefore T_(2) =250^(@)C-122.4^(@)C=127.6^(@)C` |
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| 284. |
The thermal conductivity of brick is `1.7 W m^(-1) K^(-1)`, and that of cement is `2.9 W m^(-1) K^(-1)`.What thickness of cement will have same insulation as the brick of thickness `20 cm`. |
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Answer» Since `Q= (KA(T_(1)-T_(2))t)/(L)`. For same insulation by the brick and cement `Q, A(T_(1)-T_(2))` and t do not change. Hence , `(K)/(L)` remain constant . If `K_(1)` and `K_(2)` be the thermal conductivity of brick and cement respectively `L_(1)` and `L_(2)` be the required thickness then `(K_(1))/(L_(1)) = (K_(2))/(L_(2))` or `(1.7)/(20) = (2.9)/(L_(2))` `therefore` `" "` `L_(2) = (2.9)/(1.7) xx 20 = 34.12 cm` |
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| 285. |
A concave mirror of focal length `20 cm` is cut into two parts from the middle and the two parts are moved perpendicularly by a distance `1 cm` from the previous principal axis `AB` .find the distance between the images formed by the two parts? A. 2 mmB. 6mmC. 3mmD. 4mm |
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Answer» Correct Answer - A For `M_(1): u=-10"cm"` f=-20 cm , `h_(0)= -0.1 "cm"` `v=(200)/(-10+20) = (200)/(10) implies v=20"cm"` `m=-(v)/(u) implies (h_(1))/(h_(0)) = (+20)/(+10)` `implies h_(1) = 2 xx -0.1 = -0.2 , h_(1) = -2` mm For `m_(2) = (-20)/(-10) = 2 implies h_(1) = 0.2 "cm" = 2"mm"` |
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| 286. |
The density in grams per litre of ethylene `(C_(2)H_(4))` at `STP` is :-A. `1.25`B. `2.50`C. `3.75`D. `5.25` |
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Answer» Correct Answer - A `PV = nRT implies P = (rho)/(M_(w)) RT` `implies rho = (PM_(w))/(RT)= - ((10^(5))(28 xx 10^(-3)))/(8.3 xx 273) kg m^(-3) = 1.25 g //litre` |
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| 287. |
Calculate the value of universal gas constant at `STP`. |
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Answer» Universal gas constant is given by ` R=(PV)/(T)` One mole of all gases at ` S.T.P.` occupy volume `V = 22.4` litre = ` 22.4xx10^(-3)m^(3)` `P = 760` mm of Hg = ` 760xx10^(-3)xx13.6xx 10^(3) xx9.80 N m^(-2) T=273.16 K` `therefore R = (760xx10^(-3) xx 13.6 xx 10^(3)xx 9.80 xx 22.4 xx 10^(-3))/273.16 = 8.314 J mol^(-1)K^(-1)` |
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| 288. |
By increasing temperature of gas by `5^(@)C` its pressure increase by `0.5%` from its initial value at constant volume then what is initial temperature of gas? |
| Answer» `because` At constant volume `T alpha P` `therefore (DeltaT)/(T) xx100 = (DeltaP)/(P) xx100 = 0.5 implies T=(5xx100)/(0.5) = 1000K` | |
| 289. |
Find the quantity of heat required to convert 40 g of ice at `-20^(@)`C into water at `20^(@)`C. Given `L_(ice) = 0.336 xx 10^(6) J/kg.` Specific heat of ice = 2100 J/kg-K Specific heat of water = 4200 J/kg-K |
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Answer» Heat required to raise the temperature of ice from `-20^(@)C` to `0^(@)C= 0.04xx2100xx20 =1680J` Heat required to convert the ice into water at `0^(@)C` = `mL = 0.04xx0.336xx10^(6)=13440J` Heat required to heat water from `0^(@)C` to `20^(@)C = 0.04xx4200xx20 = 3360J` Total heat required =`1680+13440+3360 = 18480J` |
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| 290. |
In figure , heat is added to a pure substance in a closed container rate. A graph of the temperature of the substance as a function of time is shown here , If `L_(F)` = heat of fusion and `L_(V)`= heat of vaporization . What is the value of the ratio `L_(V)//L_(F)` for this substance ? A. `5.0`B. `4.5`C. `7.2`D. `3.5` |
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Answer» Correct Answer - D `L prop` number of blocks |
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| 291. |
Steam at `100^(@)C` is passed into `1.1 kg` of water contained in a calorimeter of water equivalent `0.02kg` at `15^(@)C` till the temperature of the calorimeter and its content rises to `80^(@)C`. What is the mass of steam condensed? Latent heat of steam = `536 cal//g`. |
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Answer» Heat required by (calorimeter +water) `Q = (m_(1)c_(1) +m_(2)c_(2))Deltatheta = (0.02+1.1xx1)(80-15) = 72.8kcal` If m is mass of steam condensed , then heat given by steam `Q` =m`L` +mc `Deltatheta` = `m xx536 + m xx1xx(100-80)=556m` ` therefore 556m = 72.8` `therefore` Mass of steam condensed `m=72.8/556 = 0.130kg` |
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| 292. |
Steam at `100^(@)C` is passed into `1.1 kg` of water contained in a calorimeter of water equivalent `0.02kg` at `15^(@)C` till the temperature of the calorimeter and its content rises to `80^(@)C`. What is the mass of steam condensed? Latent heat of steam = `536 cal//g`.A. `0.130`B. `0.065`C. `0.260`D. `0.135` |
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Answer» Correct Answer - D Heat lost = Heat gained ` m_("steam") xx 540 = 1100 xx 1 xx (80-15)+20xx 1 xx (80-15)` `therefore ` mass of steam condensed = `0.13` kg |
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| 293. |
Objects `A` and `B` that are initially separated from each other and well isolated from their surroundings are then brought into thermal contact. Initially `T_(A) = 0^(@)C` and `T_(B) = 100^(@)C` . The specific heat of `A` is less than the specific heat of `B`. After some time, the system comes to an equilibrium state. the final temperatures are: A. `T_(A) = T_(B) gt 50^(@)C`B. `T_(A)gt T_(B)gt 50^(@)C`C. `T_(A) = T_(B) lt 50^(@)C`D. `T_(B)gtT_(A) gt 50^(@)C` |
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Answer» Correct Answer - A At equilibrium `T_(A) = T_(B)` |
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| 294. |
The temperature of an iron piece is heated from `30^@ C "to" 90^@ C`. What is the change in its temperature on the fahrenheit scale and on the kelvin scale? |
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Answer» `Delta C=90^(@)-30^(@) = 60^(@)C` Temperature difference on Fahrenheit Scale `DeltaF =(9)/(5)DeltaC =(9)/(5)(60^(@)C)=108^(@)F` Temperature difference on Kelvin Scale ` DeltaK= DeltaC = 60K` |
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| 295. |
Consider a hypothetical situation where we are comparing the properties of two crystals made of atom `A` and atom `B`. Potential energy `(U) v//s` interatomic separation (r) graph for atom `A` and atom `B` is shown in figure (i) and (ii) and respectively. Choose correct statementA. Volume of `A` and `B` expand on heatingB. Volume of `A` and `B` contract on heatingC. `A` expands on heating and `B` contracts on heatingD. `A` contracts on heating and `B` expands on heating |
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Answer» Correct Answer - C On increasing temperature (giving heat) , `U` increases. Now `r_(avg)` increases for `A` while decreases for `B`. |
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| 296. |
Calculate the difference between the two principal specific heats of 1g of helium gas at S.T.P. Given atomic weight of helium = 4 and `J = 4.186 J cal^(-1)` and `R = 8.31 J mol^(-1)K^(-1)`. |
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Answer» Gas constant for `1` g of helium , r= `(R)/(M_(w)) = (PV)/(TxxM_(w)) = (76xx13.6xx981xx22400)/(273xx4) = 2.08 xx10^(7) "erg" g^(-1) K^(-1)` `C_(P) -C_(V) = (r)/(J) = (2.08xx10^(7))/(4.186xx10^(7)) = 0.5"cal" g^(-1) K^(-1)` |
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| 297. |
When water is heated from `0^(@)C` to `4^(@)C` and `C_(P)` and `C_(V)` are its specific heats at constant pressure and constant volume respectively , then :-A. `C_(P)gt C_(V)`B. `C_(P)lt C_(V)`C. `C_(P) = C_(V)`D. `C_(P) - C_(V) = R` |
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Answer» Correct Answer - B When water is heated from `0^(@)C` to `4^(@)C` , its volume decreases. ` therefore P DeltaV` is negative Hence `C_(P) - C_(V) lt 0 implies C_(P) lt C_(V)` |
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| 298. |
Statement - `1`: An ideal gas has infinitely many molar specific heats. Statement-`2`: Specific heat is amount of heat needed to raise the temperature of `1` mole of gas by `1K`.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - D | |
| 299. |
using euipartion of energy, the specific heat `("in" jkg^(-1)K^(-1)`of aluminium at room temperature can be estimated to be (atomic weigh of aluminium=27)A. `410`B. `925`C. `25`D. `1850` |
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Answer» Correct Answer - B Energy `E = (nfRT)/(2) = (m)/(M_(0))(6RT)/(2) … (i)` (degree of freedom for Al = `6` , `M_(0)= "molecular mass of Al")` and `Q =msT`…..(ii) From (i) & (ii) `(3mRT)/(M_(0)) = msT` |
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| 300. |
A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. Find the speed of the image formed as a function of time for `tlt sqrt((2h)/g)`. Consider only the image by a single refraction. |
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Answer» Correct Answer - `(muR^(2)g t)/([(mu-1)(y_(0)-(1)/(2)g t^(2))-R]^(2))` |
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