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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.
| 1. |
What is a systematic error ? How can it be removed ? |
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Answer» Systematic error is that which is always positive or always negative. Such an error can be removed by detecting the source of error and applying the necessary correction. |
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| 2. |
The dimensional formula for angular momentum isA. `[ML^(0)L^(2)T^(-2)]`B. `[ML^(2)T^(-1)]`C. `[MLT^(-1)]`D. `[ML^(2)T^(-2)]` |
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Answer» Correct Answer - B Angular momentum `L = r xx p = r xx mv` `:.` Dimensional formula for angular momentum `=[L] [M] [LT^(-1)] = [ML^(2)T^(-1)]` |
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| 3. |
if the universe were shrunk to the size, of earth, how large would the earth be on this scale? |
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Answer» Here, size of universe `~= 10^(25)m` `"size of earth" ~= 10^7m` `"Magnification"m = ("size of earth")/("size of universe")` `=(10^7)/(10^(25)) = 10^(-18)` `:. "Apparent size of earth" = mxx "actual size"` `= 10^(-18)xx10^7` `=10^(-11)m` |
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| 4. |
The dimensional formula for conductance isA. `[M^(-1) L^(-2) T^3 A^(-2)]`B. `[M^1 L^2 T^(-3) A^2]`C. `[M^(-1)L^(-2) T^(-3) A^2]`D. `[M^(-1)L^(-2)T^3 A^2]` |
| Answer» Correct Answer - (d) | |
| 5. |
The estimated size of observable universe is of the order of……… |
| Answer» Correct Answer - `10^(26)m` | |
| 6. |
Give two examples of non dimensional veriables. |
| Answer» Strain and specific gravity. | |
| 7. |
What are the dimensions of rate of flow? |
| Answer» Rate of flow ` =(Volume)/(time) = (L^3)/(T) = [M^0 L^3 T^(-1)]` | |
| 8. |
Age of the universe is about `10^(10)` years whereas the mankind has existed for `10^6` years. How many seconds would the man have existed if age of universe were one day. |
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Answer» Here, `("Age of mankind")/("Age of universe") = (10^6yrs)/(10^(10)yrs) = (x)/(1 day)` `x = 10^(-4)day = 10^(-4)xx86400 s= 8.64s` |
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| 9. |
State the number of significant figures in the following : (a) `0.007m^2` (b) `2.64 xx10^(24) kg` (c) `0.2370 g cm^(-3)` (d) `6.320 J` (e ) `6.032 N m^(-2)` (f) `0.0006032 m^2` |
| Answer» The number of significant figures is as follows : (a) one (b) three (c ) four (d) four (e ) four (f) four | |
| 10. |
A student measures the time period of `100` ocillations of a simple pendulum four times. The data set is `90 s`, 91 s, 95 s, and 92 s`. If the minimum division in the measuring clock is `1 s`, then the reported men time should be:A. `(92 +-2)s`B. `(92 +- 5)s`C. `(92 +- 1.8)s`D. `(92 +- 3)s` |
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Answer» Correct Answer - (a) Mean value of time period `=(90 + 91 +92 +95)/(4) = 92 s` Mean absolute error `=( | 92 - 90 | + |92- 91| + |92 - 92 | + |92 - 95|)/(4)` `=( 2+1+0 +3)/(4) = 1.5 s` :. Value of time period `= (92 +- 1.5)s` As least count = 1s :. Value of time period = `(92 +- 2) s.` |
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| 11. |
Can a quantity have units, but still be dimensionless? |
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Answer» Yes. For example, angle has units (radian) but is dimensinless. |
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| 12. |
In different of systems of units, can a quantity have different dimensions? |
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Answer» No, a quantity has same dimensions in all system of units. |
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| 13. |
A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate object under water. In a submarine equaipped with as SONAR, the time delay between genration of a probe wave and the recption of its echo after refection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine ? (speed of sound in water `= 1450 ms^(-1)` |
| Answer» Here, `t = 77.0s, x = ? v = 1450 ms^(-1)`. As `x = (upsilonxxt)/(2) :. X =(1450xx77.0)/(2) m = 55825m` | |
| 14. |
The farthest objects in out universe discovered by modern astronomeres are so distant that light emitted by them takes billions of year to reach the earth. These object (known as quasers) have may puzzling features, which have yet not been satisfactorily explained. What is the distance in km of a quasar form which light takes 3.0 billion years to reach us? |
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Answer» Here, x= ? Time taken t =3.0 bilion years `=3xx10^9 yr = 3xx10^9xx365xx24xx60xx60s` vel. Of light in vacuum, `c = 3xx10^8 m//s = 3xx10^5 km//s` As distance = velocityxx time :. `X= (3xx10^5)xx3xx10^9xx365xx24xx60xx60 km = 2.84xx10^(22)km` |
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| 15. |
A great physicist of this century (P. A. M. Dirac) loved playing with numerical values of fundamental constant of nature. This led him to an instreasing observaion. Dirac found that form the basic constant of atomin physice (c,e, mass of electron mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe `(~~ 15 billion years).` Form the table of fundamental constants in this book, try to see if you too can construct this number (or any other instresting number you can think of). if its coincidence with the age of the universe ware significant, what would this imply for the constancy of fundamental constants ? |
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Answer» Trying out with basci constants of atomic physics (speed of light c, charge on electron e, mass of electron `m_e` mass of proton `m_p`) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. Once such quatity si `t = ((e^2)/(4pi "in"_0))^2xx(1)/(m_p m_e^2 c^3G)` Put `e = 1.6xx10^(-19)C, (1)/(4pi"in"_0) = 9xx10^9, c = 3xx10^8m//s and G = 6.67xx10^(-11) Nm^2kg^(-2)` `m_p 1.67xx1-^(-27)kg , m_e = 9xx10^(-31)kg` `t = (1.6xx10^(-19))^4xx(9xx10^9)^2xxx(1)/(1.67xx10^(-27)(9xx10^(-31))^2(3xx10^8)^3xx6.67xx10^(-11))` `t = 2.18xx10^(16) sec`. This time is of the order of age of universe. |
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| 16. |
Momentum per unit volume , divided by pressure represents reciprocal of velocity. Comment. |
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Answer» `True. (Momentum//Volume)/(pressure) = (MLT^(-1)//L^3)/(MLT^(-2)//L^2)` ` = TL^(-1) = (LT^(-1) = upsilon^9-1) = (1)/(upsilon)` |
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| 17. |
The dimensional forumal of Hubble constant. Is same as that of frequency. Comment. |
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Answer» You, true. Hubble constant ` = (velocity)/(distance) =(LT^(-1))/(L) = [T^(-1)]` which represents frequency. |
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| 18. |
Human heart is an inbulit clock. Comment |
| Answer» True. Becauese human heart beasts at a regular rate. | |
| 19. |
Energy density and pressure have the same dimensions. Comment. |
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Answer» True, because energy density `=("energy" )/("volume") = (ML^2T^(-2))/(L^3) = [M^1 L^(-1) T^(-2)]` and pressure `=("force")/("area") = (ML^(2)T^(-2))/L^(3)` `=[M^1L^(-1)T^(-2)]` |
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| 20. |
Given : force `= (alpha)/("density" + beta^3).` What are the dimensions of `alpha, beta ?`A. `M L^(-2) T^(-2), ML^(-1//3)`B. `M^2 L^4 T^(-2), M^(1//3) L^(-1)`C. `M^2 L^(-2) T^(-2), M^(1//3) L^(-1)`D. `M^2 L^(-2), M L^(-3)` |
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Answer» Correct Answer - (c ) `beta^3 = "density" = M^1 L^(-3) = beta =M^(1//3) L^(-1)` Also, `alpha = "force" xx "density"` `=MLT^(-2)xx M^1 L^(-3) = [M^2 L^9-2)T^(-2)]` |
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| 21. |
What are the dimensions of linear mass density ? |
| Answer» liner mass density = mass//length` = [M^1 L^(-1)]` | |
| 22. |
The dimensional formula of torque isA. `[ML^(2)T^(-2)]`B. `[MLT^(-2)]`C. `[ML^(-1)T^(-2)]`D. `[ML^(-2)T^(-2)]` |
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Answer» Correct Answer - A Torque `tau = r xx F` Dimensions of `tau =` dimension of `r xx` dimension of `F` `=[L] [MLT^(-2)] = [MLT^(2)T^(-2)]` |
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| 23. |
Calculate the number of light years in one kilometer. |
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Answer» As is known, `1ly = 9.46xx10^(15)m` `=9.46xx10^(12)km` `:. 1km = (1)/(9.46xx10^(12))ly` `1km = 1.057xx10^(-13)ly` |
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| 24. |
If ` x= at + bt^(2)` , where `x` is the distance rtravelled by the body in kilometer while `t` is the time in seconds , then find the units of `b`.A. `km//s`B. `km-s`C. `km//s^(2)`D. `km-s^(2)` |
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Answer» Correct Answer - C As `x = at +bt^(2)` According to the concept of dimensional analysis and principle of homogeneity `:.` unit of `x =` unit of `bt^(2)` `:.` unit of `b = ("unti of" x)/("unit of" t^(2)) = km//s^(2)` |
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| 25. |
In an experiment the refractive index of glass was observed to be `1.45 , 1.56 , 1.54 , 1.44 , 1.54 , and 1.53`. Calculate (a). Mean value of refractive index (b). Mean absolute error ( c ) Fractional error (d) Percentage error (e) Express the result in terms of absolute error and percentage error |
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Answer» Correct Answer - (i) `1.51 (ii) +- 0.4 ()iii) +- 0.03 (iv) +- 3% ; mu =1.51 +- 0.04 ; mu =1.51 +- 3%.` Mean value of `mu = (1.45 +1.56 +1.54 +1.44 +1.54 +1.53)/(6) =1.51` Absolute error are : `1.51 -1.45 = 0.06, 1.51 - 1.56 = - 0.05` `1.51 -1.54 = 0.03, 1.51 - 1.44 = - 0.07` `1.51 -1.54 = 0.03, 1.51 - 1.53 = - 0.02` Mean absolute error ` = +-(0.06 + 0.05 +0.03 +0.07 +0.03 + 0.02)/(6)` `= +- 0.26//6 = +- 0.04` `Fractional error = (+- 0.4)/(1.51) =+- 3%` `mu =1.51 +- 0.04 or mu= 1.51 +- 3%` |
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| 26. |
Two spherometers A and B have the same pitch. A has 100 division on periphery of its circular disc and B has 200 division on periphery of its circular disc. ThenA. Both A and B have same least countB. L.C. of A is twice the L.C. of BC. L.C. of A is half the L.C. of BD. Nothing can be said |
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Answer» Correct Answer - (b) As `L.C. = ("pitch")/("no. of divs. On C.S")` and B has twice the number of division on A, :. L.C. of A is twice the L.C. of B |
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| 27. |
Calculate the solid angle subtended by the periphery of an area of `1 cm^2` at a point situated symmetrically at a distance of 5 cm from the area. |
| Answer» `d omega = (ds)/(r^2) = (1cm^2)/((5cm)^2) = 4xx10^(-2) steradian.` | |
| 28. |
A drop of olive oil of radius 0.3 mm spreads into a rectangular film of `30 cmxx15cm` on the water surface. Calculate the size of the oil molecule. |
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Answer» Correct Answer - `2.51xx10^(-9)m` `d = ("Volume of drop")/("Area of film") = ((4)/(3)pir^3)/(ixxb)` `=(4)/(3)xx(22)/(7) (0.3xx10^(-3))^3/(30xx15xx10^(-4)) = 2.51xx10^(-9)m` |
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| 29. |
A drop of olive oil of diameter `0.6xx10^(-3)m` spreads into a circular film of raius 12 cm. Estimate the molecular size of olive oil. |
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Answer» Here, radius of oil drop `= r = (0.6xx10^(-3))/(2) m = 0.3xx10^(-3)m` `R =12cm =12xx10^(-2)m` Thickness of oil film ` = ("Volume of oil drop")/("Area of film") = ((4)/(3)pir^3)/(piR^3)` t=(4/3pi(0.3xx10^(-3))^(3))/(piR^(2)) = (4(0.3xx10^(3))^(3))/(3xx(12x10^(-2))^2) = (4xx0.027xx10^(-5))/(3xx12xx12) m = 2.5xx10^(-9)m`. If we assume that the film is one molecule thick, then molecular size of oleic acid `~= 2.5xx10^(-9)m` |
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| 30. |
Why length, mass and time are chosen as base quantities in mechanics ? |
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Answer» In mechanics, length, mass and time are chosen as the base quantities because (i) there is nothing simpler to length, mass and time. (ii) all other quantites in mechanics can be expressed in terms of length, mass and time. (iii) length, mass and time cannot be derived from one another. |
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| 31. |
Add 0.4382 to 876 and wirte the result with correct number of significant figures. |
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Answer» `876 + 0.4382 = 876.4382 rarr 876,` rounding off to no decimal point. |
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| 32. |
Add `2.384xx10^(-4) to 1.7xx10^(-5)` and express the result to correct number of significant figures. |
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Answer» Correct Answer - `1.9xx10^5` `2.384xx10^(-4)+1.7xx10^(-5)` `=10^(-5)[0.2384 +1.7] = 1.9384xx10^(-5)` Rounding off to one place of decimal : `sum = 1.9xx10^(-5)` |
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| 33. |
When we add 0.9825 ot 3.04, the correct result with regard to significant figures isA. 4.0225B. 4.022C. 4.02D. 4 |
| Answer» Correct Answer - (c ) | |
| 34. |
(a).Add `3.8 xx 10^(-6) to 4.2 xx 10^(-5)` with due regard to significant figures. (b). Subtract `3.2 xx 10^(-6) from 4.7 xx 10^(-4)` with regard to significant figures. ( c ). Subtract `1.5 xx 10^(3) from 4.8 xx 10^(4)` with due regard to significant figures. |
| Answer» Correct Answer - `4.7xx10^^(-4)` | |
| 35. |
Add 8.2 and 10.163 and round off the sum to proper number of significant figures. |
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Answer» 8.2 + 10.163 = 18.363 Rounding off to one decimal place, sum = 18.4 |
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| 36. |
In the equation y= a sin `(omega t+ kx)` t and x stand for time and distance respectively. What are the dimensions of `omega//k`? |
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Answer» Correct Answer - `LT^(-1)` The given equation is `y =a sin (omega t - kx)` As angle is dimensionless, therefore, `(omegat- kx)` = angle `omega =(1)/(t) = T^(-1) and k = (1)/(k) = L^(-1)` `(omega)/(k) = (T^(-1))/(L^(-1)) = [LT^(-1]` |
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| 37. |
Assertion: The error in the measurement of radius of sphere is `0.3%`. The permissible error in its surface area is `0.6 %`. Reason: The permissible error is calculated by the formula `(Delta A)/(A) = (4 Delta r)/( r)`.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 - (c ) Here, assertion is true, but reason is false. as `A = 4 pi r^2, :. (DeltaA)/(A) = (2 Delta r)/(r ) = 2xx 0.3 % = 0.6 %` Also `(DeltaA)/(A) != (4 Delta r)/(r )`, So reason is false. |
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| 38. |
The dimensions of `(a)/(b)` in the relatin F = ax + bt areA. `LT^(-1)`B. `L^(-1)T`C. LTD. `L^(-1)T^(-1)` |
| Answer» Correct Answer - (b) | |
| 39. |
Write the dimensions of `a//b` in the relation `F = a sqrtx + bt^2` where F is force x is distance and t is time. |
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Answer» Correct Answer - `[L^(-1//2) T^2]` `asqrtx` has the dimensions of force F `:. a = F|sqrtx = MLT^(-2) / L^(1//2) = M^1 L^(1//2) T^(-2)` Similarly, `b = F|t^2 = MLT^(-2)|T^2 = M^1L^1T^(-4)` `:. (a)/(b) = (ML^(1//2)T^(-2))/(MLT^(-4)) = [L^(-1//2T^2)]` |
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| 40. |
Give an example of (a) a physical quantiy which has a unit but no dimensions. (b) a physical quantity which has neither unit no dimensions. (c ) a constant which has a unit. (d) a constant which has no unit. |
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Answer» (a) Angle in a plane in measured in radian, though it has no dimensions. (b) Strain has neither units nor dimensions. (c ) Universal gravitational constant (G) has a constant value : `6.67xx10^(-11) Nm^2 kg^(-2).` (d) Reynold number is a constant which has no units. |
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| 41. |
The position x of a partical at time t is given by `x =(V_0)/(a) (1 -e^(9-at))` where `V_0` is a constant and a gt 0. The dimensions of `V_0` and a are.A. `M^0 L T^(-1) and T^(-1)`B. `M^0 L T^0 and T^(-1)`C. `M^0 L T^(-1) and L T^(-2)`D. `M^0 L T^(-1) and T` |
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Answer» Correct Answer - (a) As axxt is dimensionless. `:. a = (1)/(t) = (1)/(T) = [T^(-1)]` Also,` x = (V_0)/(a), V_0 = x a = [LT^(-1)]` |
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| 42. |
Find the dimensios of `a//b` in the relation `P = ax +bt^2,` where P is pressure, x is distance and t is time. |
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Answer» As ax = p, `a = (P)/(x) = (ML^(-1)T^(-2))/(L) = [ML^(-2)T^(-2)]` Again, as `bt^2 = P,` `b =(P)/(t^2) = (ML^(-1)T^(-2))/(T^2) = [ML^(-1)T^(-4)]` `(a)/(b) = (ML^(-2)T^(-2))/(ML^(-1)T^(-4)) = [M^0L^(-1)T^2]` |
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| 43. |
The dimensions of T in the dimensional formula for mobilitty are : |
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Answer» Correct Answer - 2 Mobility `= ("drift velocity")/("electric field") = (upsilon)/(F//q) = (q upsilon)/(F)` `((ATLT^(-1)))/((MLT^(-2))) = [M^(-1) L^0 T^2 A^1]` The dimension of T =2 |
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| 44. |
Assertion : Distance travelled by the particle in the nth second has dimensios of length. Reason :It is the distance travelled by the partical in the given time.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) Here, both the assertion and reason are false. `As S_(nth) = u +(a)/(2) (2n-1)` `:. S_(nth) = LT^(-1) + LT^(-2). T =[LT^(-1)]` Also, it is the distance travelled by particle in particular second, not in the given time. |
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| 45. |
The diameter of a wire as measured by a screw gauge was found to be 0.026 cm, 0.028 cm, 0.029 cm, 0.027cm, 0.024cm and 0.027 cm. Calculate (i) mean value of diameter (ii) mean absoulte error (iii) relative error (iv) percentage error. Also express the result in terms of absolute error and percentage error. |
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Answer» Correct Answer - (i) 0.027cm 9ii) 0.001 cm(iii) 0.037 (iv) 3.7 % (0.027 +- 0.001) cm 0.027 cm +- 3.7 %` (i) Mean value of diameter, `D = (0.026 + 0.028 + 0.029 +0.027 +0.024 +0.027)/(6)` `D = (0.161)/(6) = 0.268 =0.027cm` Now, `Delta D_1 =0.027 - 0.026 =0.001` `Delta D_2 =0.027 - 0.028 =0.002` `Delta D_3 =0.027 - 0.029 =0.002` `Delta D_4 =0.027 - 0.027 =0.000` `Delta D_5 =0.027 - 0.024 =0.003` `Delta D_6 =0.027 - 0.027 =0.000` (ii) Mean absolute arror `Delta D_1 =0.027 - 0.026 =0.001` `=(| DeltaD_1 |+| Delta D_2 | + |Delta D_3 + |/Delta D_4| + | DeltaD_5|+|DeltaD_6|)/(6)` `Delta D_1 =0.027 - 0.026 =0.001` `= (0.001+ 0.001 + 0.002 + 0.000 +0.003 +0.000)/(6)` `Delta D_1 =0.027 - 0.026 =0.001` `=(0.007)/(6) = 0.001cm` (iii) Relative error `= (0.001)/(0.027) = 0.037` (iv) Percentage error `= 0.037xx10% = 3.7%` (v) Diameter of wire `= (0.27 +- 0.001)cm` `=0.027cm +- 3.7%` |
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| 46. |
Assertion: The dimensional formula of surface energy is `[M^(0)L^(2)T^(-2)]`. Reason: surface energy has same dimensions as that of potential energy.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) Both the statements are true and statement -2 is correct explanation of statement-1. |
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| 47. |
Statement -1 : Distance travelled in nth second has the dimensions of velocity. Statement -2 : Because it is the distancce travelled in one (particular) second.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) Both the statements are true and statement -2 is correct explanation of statement-1. |
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| 48. |
If force `(F)` velocity `(V)` and time `(T)` are taken as fundamental units, then the dimensions of mass areA. `[FVT^(-1)]`B. `[FVT^(-2)]`C. `[FV^(-1) T^(-1)]`D. `[FV^(-1)T]` |
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Answer» Correct Answer - (d) Dimensional formula of force `F = [MLT^(-2)]` or `F =[M(LT^9-1)T^(-1)] = [(MVT^9-1)]` or `M = [FV^(-1) T]` |
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| 49. |
If the speed of light c, acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities then the dimension of gravitational constant isA. `c^0 g p^(-3)`B. `c^2 g^3 p^(-2)`C. `c^0 g^2 p^(-1)`D. `c^2 g^2 p^(-2)` |
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Answer» Correct Answer - (c ) Let `G = c^x g^y p^z` `[M^(-1)L^3 T^(-2)] = (LT^(-1 ^(x))) (LT^(-2))^y (ML^(-1) T^(-2))^z` `=M^z L^(x+y) T^(-x-2y-2z)` Applying principle of homogeneity of dimensions, we get z = 1, x+y - x=3. -x -2y -2z = -2 On solving, we get, y= 2, z=0 `:. G = c^0 g^2 p^(-1)` |
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| 50. |
In terms of basic units of mass (M), length (L), time (T), and charge (Q), the dimensions of magnetic permeability of vacuum `(mu_0)` would beA. `(MLQ^(-2))`B. `(LT^(-1) Q^(-1))`C. `(ML^2 T^(-1)Q^(-2))`D. `(LTQ^(-1))` |
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Answer» Correct Answer - (a) Force acting per unit length of eacth of two parallel wires carrying currents in the same direction placed certain distance apart in vacuum is `(F)/(I) = (mu_0)/(4pi). (2I_1II_2)/(r )` `(MLT^(-2))/(L) = mu_0 ((Q//T)^2)/(L) or mu_0 = [MLQ^(-2)]` |
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