InterviewSolution
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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.
| 251. |
Which is the smalles prectical unit of time ? |
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Answer» The smallest practical unit of time is `1 shake = 10^(-8)s.` |
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| 252. |
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 division is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw are brought in constant, 45th division coincides with the main scale line and that the zero of main scale line and that the zero of main scale is barely visible. What is the thickness of the sheet, if the main scale reading is 0.5 mm and 25th division coincides with the main scalel line ?A. 0.75 mmB. 0.80mmC. 0.70 mmD. 0.50mm |
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Answer» Correct Answer - (b) Here, pitch = 0.5 mm Least count `= (0.5 mm)/(50) = 0.01mm` Zero error = `-(50 -45)xx0.01 = - 0.05mm` Zero correction `= + 0.05 mm` Thickness of sheet `= 0.5 mm + 25 xx 0.01 mm` + `0.05mm 0.80 mm` |
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| 253. |
The screw of a spherometer moves by 4 mm. when its circular scale is given four complete rotations. If circular scale has 200 divisions, calculate pitch and least count of the spherometer. |
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Answer» `Pitch = ("distacne moved")/("number of complete rotations")` `(4mm)/(4) = 1mm` Least count = `("Pitch")/("no. fo divisions on circular scale")` `=(1mm)/(200) = 0.005 mm = 0.0005 cm` |
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| 254. |
When circular scale of a screw gauge carrying 100 divisions is given four complete rotation, the head of the screw moves through 2mm. Calculate pitch and least count of the screw gauge. |
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Answer» Here, no of divisions on circular scale =100 no. of complete rotations = 4 distacne moved =2mm. ` Pitch = ("distance moved")/("no. of complete rotations") = (2mm)/(4)` 0.5mm Least count =` ("Pitch")/("no. ofn divisions on circular scale")` `=(0.5mm)/(100) = 0.005mm` |
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| 255. |
Consider a Vernier callipers in which each `1cm` on the main scale is divided into `8` equal divisions and a screw gauge with `100` divisions in its circular scale. In the Vernier callipers, `5` divisions of the Vernier scale coincide with `4` divisions on the scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then :A. If the pitch of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.01 mm.B. If the patich of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.005 mm.C. If the least conunt of the linear scale of the screw gauge is twice the least count of screw gauge is 0.01 mm.D. If the least count of the linear scale of the screw gauge is twice the least count of the vernier callipers, the least count of screw gauge is 0.005mm. |
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Answer» Correct Answer - (b,c) Smallest division on main scale of vernier callipers `=(1)/(8) cm = 0.125 cm` 5 division of the vernier scale = 4 div. of main scale = 4xx 0.125 = 0.5 cm 1 division of vernier scale `=(0.5)/(5) = 0.1 cm` L.C of vernier callipers = 1 MSD - 1 VSD =0.125 - 0.1 = 0.025 cm L.C. of screw gauge `= (pitch)/(100)` If pitch of screw gauge is twice the L.C of vernier callipers, then L.C of screw gauge `=(0.025xx2)/(100) = (0.05)/(100)cm = (0.5)/(100) mm` =0.05 mm Hence option (b) si correct. Also ,If L.C of linear scale of screw gauge is twice of L.C. of vernier callipers, then L.C of linear scale of screw gauge =2xx0.025 = 0.05 cm pitch = 2xx 0.05 = 0.1 cm = 1mm L.C of screw gauge `=(1mm)/(100) = 0.01mm` Hence option (c ) is correct. |
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| 256. |
Which of the following is the most precise devise for measuring length ? (a) a Vernier callipers with 20 divisions on the sliding scals, coindiing with 19 main scale divions (b) a screw gauge of pitch 1mm and 100 divisions on the circular scale (c ) an optical instrument that can measure length to within a wave length of light. |
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Answer» The most precise device is that whose least count is minimum. Now (a) Least count of theis vernier callipers ` = 1 SD -1VD = 1SD -(19)/(20)SD = (1)/(20)SD` `=(1)/(20)mm = (1)/(200) = 0.005cm` (b) Least count of screw gauge ` = ("pitch")/("no. of divisonsn on circular scale") = (1)/(100)mm = (1)/(1000) cm = 0.001cm`. (c ) Wavelength of light,`lambda ~~ 10^(-5)cm. =0.00001cm`. Obviousoly, the most precise measurement is with optical instrument. |
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| 257. |
Finding dimensions of resistance R and inductance L, speculate what physical quantities `(L//R) and (1)/(2)LI^2` represent, where I is current? |
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Answer» The dimensions of R and L are : `R = [M^1L^2 T^(-3)A^(-2)]` and `L= [ML^2 T^(-2)A^(-2)]` Now, `(L)/(R ) = (ML^2T^(-2) A^(-2))/(ML^2 T^(-3)A^(-2)) = T^1` it represents time constant of RL cirucit. Again `(1)/(2)LI^2 = [ML^2 T^([email protected])A^(-2)]xxA^2` `=[M^1 L^2 T^(-1)]` it represents (magnetic) energy stored in an inductor. |
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| 258. |
What efforts have been made towards unification of forces ? |
| Answer» Correct Answer - 1 (a) 10 | |
| 259. |
The refractive index `mu` of a medium is found to vary with wavelength `lambda` as `mu = A +(B)/(lambda^2).` What are the dimensions of A and B? |
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Answer» mu = `("velocity of light in vacuum")/("velocity of light in medium")` `= "dimensionless number" ` A`mu = A +(B)/(lambda^2),therefore`, A must be dimensionless, and `B = mu lambda^2 = [L^2]` |
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| 260. |
Deduce the dimensional formula of thermal conducitvity (k). |
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Answer» The rate at which heat is conducted through a solid rod is `(DeltaQ)/(Deltat) = kA((DeltaT)/(Deltax)) …..(i)` where A is area of cross section, `DeltaT//Deltax` temperature gradient, `DeltaQ` is heat conducted in `Deltat` second , k is coefficient of thermal conducitvity. From (i) `k= (DeltaQ)/(Deltat)xx(1)/(A) ((Deltax)/(DeltaT))` `=(ML^2 T^(-2))/(T)xx(1)/(L^2) (L)/(K)` `k =[MLT^(-3) k^(-1)]` Here, K represents temp. `Delta T` in kelvin. |
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| 261. |
Why is platinum iridium alloy used in making prototype metre and kilogram ? |
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Answer» This is becuause this alloy (i) is least affected by temperature Variations (ii) is non corrosive, does not ware out easily (iii) does not change with time. |
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| 262. |
Among which type of elementary particles does the electromagnetic force act ? |
| Answer» Electromagnetic force acts on all electrically charged particles. | |
| 263. |
What are conserved quantities in nature ? Name any two. |
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Answer» The physical quantities which remain unchanged in a process are called conserved quantities. For example : linear momentum and energy. |
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| 264. |
Statement-1 : Velocity gradient has the dimensions of frequency. Statement -2 : Velocity gradient is rate of change of velocity with distance.A. Statement -1 is true, Statement -2 is true , and Statement -2 is correct explanation of Statement -1.B. Statement -1 is true , Statement -2 is true, but Statement -2 is not a correct explanation of Statement -1.C. Statement-1 is true, but Statement -2 is false.D. Statement-1 is false, but Statement -2 is true. |
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Answer» Correct Answer - (a) Velocity gradient = change of velocity with distance `=(LT^(-1))/(L) = [T^(-1)] and T^(-1)` represents frequency. Therefore, both the statements are true and statement -2 is correct explanation of the statement-1. |
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| 265. |
Assertion : Pressure can be subracted from pressure gradient. Reason : Because both have the same dimensions.A. If both 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 a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - (d) Pressure gradient `=(dP)/(dx) = (ML^(-1)T^(-2))/(L)` `=M^1 L^(-2) T^(-2),` which are not the dimensions of pressure. Therefore, pressure cannot be subtracted from pressure gradient. |
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| 266. |
If `E , M , J , and G` , respectively , denote energy , mass , angular momentum , and gravitational constant , then `EJ^(2) //M^(5) G^(2)` has the dimensions ofA. LengthB. MassC. TimeD. Angle |
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Answer» Correct Answer - (d) Here, `E = Energy (M^1 L^2 T^(-2)), m = mass (M^1)` L = angular momentum `= m upsilon r =(ML^2 T^(-1))` G = gravitational constant `= (M^(-1) L^3 T^(-2))` `:. (EL^2)/(m^5 G^2) = M^1 L^2 T^(-2)xx(ML^2 T^(-1))^2/(M^5 (M^(-1)L^3 T^(-2))^2)` `=(M^3 L^6 T^(-4))/(M^3 L^6 T^(-4)) = M^0 L^0 T^0` which are the dimensios of Angle `=(arc)/(radius) = M^0 L^0 T^0` |
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| 267. |
Round off to three significant digits (i) 0.03927kg (ii) `4.085xx10^8s` |
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Answer» Correct Answer - `0.393kg; (ii) 4.08xx10^8 s` (i) 0.0393kg (:. The last digit 7gt5) (ii)`4.08xx10^8s` (:. The digit to be rounded off is 5 and percending digit is even) |
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| 268. |
Round off the following numbers as indicated : (ii) 25.653 ot 3 digits (ii) `4.996xx10^5 to 3 digits` (iii) 0.6995 to 1 digit (iv) 3.350 to 2 digits. (v) 3.450 to 2 digits. |
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Answer» (i) 25.7 (ii) `5.00xx10^5` (iii)0.7 (iv) 3.4 (v) 3.4. |
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| 269. |
The nearest star to our solar system is 4.29 light years away. How mcuh is this distance in terms of par sec ? How mcuh parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun? |
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Answer» `x = 4.29 ly = 4.29xx9.46xx10^(15)m = (4.29xx9.46x10^(15))/(3.08xx10^(16)) par sec = 1.323 par sec` `theta = (I)/(r ) = (2AU)/(x) = (2xx1.496xx10^(11))/(4.29xx9.46xx10^(15)) radian = 1.512sec` |
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| 270. |
Write in ascending order : light year, astronomical unit, par sec. |
| Answer» Astronomical unit , light year , par sec. | |
| 271. |
The average wavelength of light from a sodium lamp is `5893 Å.` Express it in (i) metre (ii) nanometer. |
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Answer» Here`lambda = 5893 Å` (i) As `1 Å = 10^(10)m, :. lambda = 5893xx10^(10)m.` As `1Å =10^(10)m = (1)/(10)xx10^(-9)m = (1)/(10)nm` `:. lambda =5893xx(1)/(10)nm = 589.3 nm` |
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| 272. |
Calculate the number of astronomical units in one metre. |
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Answer» We know, `1AU =1.5xx10^(11)m or 1.5xx10^(11)m = 1AU` `:. 1m =(1)/(1.5xx10^(11)) AU =6.67xx10^(-12)AU` |
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| 273. |
7.893 gram of a substance occupies a volume of `1.1cm^3` The density of substance with appropriate significant figures isA. `7.175g cm^(-3)`B. `7.2 g cm^(-3)`C. `7.18g cm^(-3)`D. `7.1754 g cm^(-3)` |
| Answer» Correct Answer - (b) | |