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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Why do the pendulum clocks go slow in summer and fast in winter? |
| Answer» The length of the pendulum used in clocks increases in summer and hence `T` increases whereas in winter, the length of the pendulum decreases, so `T` decrease. `T` increases means clock goes slow. | |
| 2. |
Why is the second resonance found feebler than the first ? |
| Answer» In the case of second resonance, energy gets distributed over a larger region and as such second resonance becomes feebler. | |
| 3. |
The pitch of a screw gauge is `1 mm and there are 100 divisions` on the circular scale. In measuring the diameter of a sphere there are six divisions on the linear scale and forty divisions on circular scale coincide with the reference line. Find the diameter of the sphere. |
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Answer» `LC = (1)/(100) =0.01 mm` Linear scale reading ` =6("pitch") = 6mm` Circular scale reading `=n(LC) = 40xx 0.01=0.4 mm` Total reading `=(6 +0.4)=6.4 mm`. |
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| 4. |
The distance moved by the screw of a screw gauge is `2 mm` in four rotations and there are `50` divisions on its cap. When nothing is put between its jaws, `20th` divisions of circular scale coincides with reference line, and zero of linear scale is hidden from circular scale when two jaws touch each other or zero circular scale is laying above the reference line. When plate is placed between the jaws, main scale reads `2` divisions and circular scale reads `20` divisions. Thickness of plate is.A. 1.1 mmB. 1.2 mmC. 1.4 mmD. 1.5 mm |
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Answer» Correct Answer - D Distance moved in one rotation = `0.5 mm` Least count, `LC = (0.5 mm)/(50 "divisions")=0.01 mm` Screw gauge has negative zero error. This error is `(50 - 20) 0.01 mm` or `(30) (0.01) mm`. Thickness of plate = `(2 xx 0.5 mm)+(30 +20)(0.01) mm` =` 1.5 mm`. |
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| 5. |
The pitch of a screw gauge is `1 mm` and three are `100` divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies `6` divisions below the reference line. When a wire a placed between the jaws, `2` linear scale divisions are clearly visible while ` 62` divisions on circular scale coincide with the reference line. Determine the diameter of the wire. |
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Answer» Correct Answer - B `LC = (1 mm)/(100) = 0.01 mm` The instrument has a positive zero error, `e = + n(LC) = +(6 xx 0.01) =0.06 mm` Linear scale reading = `2 xx (1 mm) = 2 mm` Circular scale reading = `62 xx (0.01 mm) = 0.62 mm` `:.` Measured reading = `2 + 0.62 = 2.62 mm` or true reading = `2.62 - 0.06` = `2.56 mm`. |
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| 6. |
A vernier callipers having `1` main scale division `= 0.1 cm` to have a least count of `0.02 cm`.If `n` be the number of divisions on vernier scale and `m` be the length of vernier scale, then.A. `n =10, m =0.5 cm`.B. ` n = 9, m = 0.4 cm`.C. ` n = 10,m = 0.8 cm`.D. ` n = 10, m = 0.2 cm`. |
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Answer» Correct Answer - C `1 VSD = (0.8 cm)/(10) = 0.08 cm` `1 MSD = 0.1 cm` `:. LC = 1 MSD - 1 VSD` =`0.1 cm - 0.08 cm` = `0.02 cm`. |
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| 7. |
The main scale of a vernier callipers reads `10 mm` in `10` divisions. Ten divisions of vernier scale coincide with nine divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with `9` main scale divisions and zero of the vernier is to the right of zero of main scale, when a cylinder is tighty placed between the two jaws, the zero of the vernier scale lies slighty to the left of `3.2 cm` and the fourth vernier division coincides with a main scale division. Find diameter of the cylinder. |
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Answer» Correct Answer - A::C `LC = ("Smallest division on main scale")/("Number of divisions on vernier scale")` `=(1mm)/(10)=0.1mm =0.01cm` Positive zero error = `N +x(LC)` =`0 + 5 xx 0.01` =`0.05 cm` Diameter =`3.2 +4 xx 0.01 = 3.24 cm` Actual diameter =`3.24 -0.05 =3.19 cm` . |
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| 8. |
In a vernier callipers, `N` divisions of the main scale coincide with `N + m` divisions of the vernier scale. what is the value of `m` for which the instrument has minimum least count. |
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Answer» Correct Answer - A `(N +m)VSD =(N) MSD` `rArr 1 VSD =((N)/(N+m))MSD` lt brgt `LC = 1 MSD -1 VSD= 1MSD-((N)/(N+m))MSD` =`m/(N+m)MSD=((1)/(1+N//m))MSD` Now, least count will be minimum for m = 1. |
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| 9. |
In the previous question, minimum possible error in area measurement can be.A. `+- 0.02 cm^2`B. `+- 0.01 cm^2`.C. `+- 0.03 cm^2`.D. zero |
| Answer» Correct Answer - D | |
| 10. |
Corresponding to given observation calculate speed of sound. Frequency of tuning fork =340 Hz. `{:("Resonance","Length from the water level in (cm)",),(,"During falling","During rising"),("First",23.9,24.1),("Second",73.9,74.1):}`. |
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Answer» Correct Answer - C::D Mean length from the water level in first resonance is `l_1=(23.9 +24.1)/2` =`24.0 cm` Similarly, mean length from the water level in second resonance is `l_2=(73.9 +74.1)/2` `= 74.0 cm` `:.` Speed of sound, `v=2f(l_2-l_1)` =`2xx340(0.740-0.240)` =`340 m//s`. |
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| 11. |
Vernier constant is the (One or more than one correct option may be correct option may be correct):A. value of one MSD divided by total nimber of divisions on the main scale.B. value of one VSD divided by total number of divisions on the vernier scale.C. total number of divisions on the main scale divided by total number of divisions on the vernier scale.D. difference between the value of one main scale division and one vernier scale division. |
| Answer» Correct Answer - D | |
| 12. |
To find index error `e` distance between object needle and poler of the concave mirror is `20 cm`. The separation between the indices of object needle and mirror was observed to be `20.2 cm` in some observation, the observed image distance is `20.2 cm` and the object distance is `30.2 cm` find (a) the index error e. (b) focal length of the mirror `f`. |
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Answer» Correct Answer - A::B::C (a) index error e =observed distance - actual distance. = separation between indices- distance between object needle and pole of the mirror `20.0- 20.0= 0.2 cm` (b) `|u|= 30.2 -0.2=30 cm` `:. u =-30 cm` `|v| = 20.2 -0.2=20 cm` `:. v = - 20 cm` Using the mirror formula, `1/f=1/v+1/u=1/(-20)+1/(-30)` or `f =-12 cm`. |
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| 13. |
In `u-v` method to find focal length of a concave mirror, if object distance is found to be `10.0 cm` and image distance was also found to be `10.0 cm`, then find maximum permissible error in (f). |
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Answer» Correct Answer - C Using the mirror formula, `1/v+1/u=1/f` we have, `1/(-10)+1/(-10)=1/f` `rArr f=-5` or `|f|=5 cm` Now, differentiating Eq. (i). we have, `(-df)/(f^2)=-(du)/(u^2)-(dv)/(v^2)` This equation can be written as `|Deltaf|_(max)=[(|Deltau|)/(u^2)+(|Deltav|)/(v^2)](f^2)` Substituting the values we get, `|Deltaf|_(max)=[(0.1)/((10)^2)+(0.1)/((10)^2)](5)^2=0.05 cm` `:. |f|=(5+- 0.05)cm`. |
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| 14. |
A student performed the experiment of determination of focal length of a concave mirrior by (u-v) method using ac optical bench of length `1.5m`. The focal length of the mirrir used is `24 cm`. The maximum error in the location of the image can be ` 0.2 cm`. The 5 sets of (u,v) values recorded by the student (in cm) are `(42,56), (48,48),(60,33),(78,39)`. the data set (s) that cannot caome from experiment and is (are) incorrectly recorded, is (are) (a) `(42,56)` `(48,48)` `(66,33)` `(78,39)`. |
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Answer» Values of options ( c) and (d) do not match with the mirror formula, `1/v+1/u=1/f`. |
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| 15. |
Select the incorrect statement.A. If the zero of vernier scale does not coincide with the zero of the main scale, then the vernier callipers is said to be having zero error.B. Zero correction has a magnitude equal to zero error but sign is opposite yo that of zero error.C. Zero error is positive when the zero of vernier scale lies to the left of the zero of the main scale.D. Zero error is negative when the zero of vernier scale lies to the zero of the main scale. |
| Answer» Correct Answer - C | |
| 16. |
Least count of screw gauge is defined as.A. `("distance moved by thimble on main scale")/("number of rotation of thimble")`B. `("pitch of the screw") /(" number of divisions on circular scale")`C. `("number of rotation of thimble") /(" number of circular scale divisions")`D. None of the above. |
| Answer» Correct Answer - B | |
| 17. |
For positive error, the correction is.A. PositiveB. negativeC. nilD. may be positive or negative |
| Answer» Correct Answer - B | |
| 18. |
The given diagram represents a screw gauge. The circular scale is divided into `50` divisions and the linear scale is divided into millimeters. If the screw advances by `1 mm` when the circular scale makes `2` complete revolutions, find the least count of the instrument and the reading of the instrument in the figure. . |
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Answer» Correct Answer - A::B::C Pitch of the screw = `(1 mm)/ 2` = `0.5 mm` Least count = `(0.5)/(50)` = `0.01 mm` Reading = Linear scale reading + (coinciding circular scale `xx` least count) = `3.0 mm +(32 xx 0.01) = 3.0 + 0.32` = `3.32 mm`. |
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| 19. |
In a metre bridge, the gaps are closed by two resistance `P` and `Q` and the balance point is obtained at `40 cm`. When `Q` is shunted by a resistance of `10 Omega`, the balance point shifts to `50 cm`. The values of `P` and `Q` are. .A. ` 10/3 Omega, 5 Omega`B. `20 Omega, 30 Omega`C. `10 Omega, 15 Omega`D. ` 5 Omega, 15/2 Omega` |
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Answer» Correct Answer - A For meter bridge to be balanced `P/Q =( 40)/(60) = 2/3` `:. p =2/3Q` When `Q` is shunted, i.e. a resistance of `10 Omega` is connected in parallel across `Q`, the net resistance becomes `(10 Q)/(10 + Q)` Now, the balance point shifts to `50 cm`, i.e. `p/(((10Q)/(10 + Q))) = 1` `:. 2/3 = (10)/(10 + Q)` `:. 20 + 2Q = 30` `:. Q = 5 Omega` and `P = (10)/(3) Omega`. |
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| 20. |
When `0.2 kg` of brass at `100 .^(@) C` is dropped into `0.5 kg` of water at `20 .^(@) C`,the resulting temperature is `23 .^(@) C`. The specific heat of brass is.A. `0.41 xx 10^3 Jkg^-1 .^(@) C^-1`.B. `0.41 xx 10^2 Jkg^-1 .^(@) C^-1`.C. `0.41 xx 10^4 Jkg^-1 .^(@) C^-1`.D. `0.41 Jkg^-1 .^(@) C^-1`. |
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Answer» Correct Answer - A Heat lost = Heat gained `:. m_1s_1DeltaT_1 = m_2s_2DeltaT_2` `:. s_1 = (m_2s_2DeltaT_2)/(m_1DeltaT_1)` = `(0.5 xx 4.2 xx 10^3 xx 3)/(0.2 xx 77)Jkg^-1 .^(@) C^-1` `:. s_1 = 0.41 xx 10^3 Jkg^-1 .^(@) C^-1`. |
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| 21. |
In a certain observation we get `l =23.2 cm,r=1.32cm` and time taken for 20 oscillations was 20.0 sec. Taking `pi^2=10,`find the value of g in proper significant figures. |
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Answer» Correct Answer - B Equivalent length of pendulum, `L =23.2 cm+1.32 cm=24.52cm` =`24.5 cm` (according to addition rule of significant figures) Time period,` T=(20.0)/(20)=1.00 s`. Time period has `3` significant figures Now, `g = (4pi^2)l/(T^2)=(4xx10xx24.5xx10^-2)/(1.00)^2=9.80 m//s^2`. |
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| 22. |
The mass of a copper calorimerter is `40 g` and its specific heat in (SI) units is `4.2 xx 10^2 J kg^-1 .^(@) C^1` The thermal capacity is.A. `4 J ^(@)C^-1`B. `18.6 J`C. `16.8 j//kg`D. `16.8 J^(@) C^-1` |
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Answer» Correct Answer - D Thermal capacity = ms =`(0.04 kg)(4.2 xx 10^2 J kg^-1 .^(@) C^-1)` =` 16.8 j// .^(@) C`. |
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| 23. |
The mass, specific heat capacity and the temperature of a solid are `1000g,(1)/(2) cal/g - .^(@)C` and `80^(@)C` respectively. The mass of the liquid and the calorimeter are `900g` and `200g`. Initially,both are at room temperature `20^(@)C` Both calorimeter and the solid are made of same material. In the steady state, temperature of mixture is `40^(@)C`, then specific heat capacity of the unknown liquid. |
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Answer» Correct Answer - A::C `m_1`=mass of solid =`1000g,S_1=`Specific heat of solid`=1/2 cal//g-.^(@)C` `=S_2` or specific heat of calorimeter `m_2` = mass of calorimeter `=200g` `m_3` =mass of unknown liquid `= 900g` `S_3` =specific heat of unknown liquid from law of heat exchange, Heat given by solid= Heat taken by calorimeter + Heat taken by unknown liquid `:. m_(1)S_(1)|Delta theta_(1)|=m_(2)S_(2)|Delta theta_(2)|+m_(3)S_(3)|Delta theta_(3)|` `:. 1000xx (1)/(2)xx (80-40)=200xx1/2(40-20)+ 900 xxS_(3)(40-20)` Solving this equation we get, `S_3=1 cal//g-.^(@) C`. |
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| 24. |
The smallest division on main scale of a vernier caplliers is `1 mm and 10` vernier divisions coincide with `9` main scale divisions. While measuring the length of a line, the zero mark of the vernier scale lies between `10.2 cm and 10.3 cm` and the third division of vernier scale coincides with a main scale division. (a) Determine the least count of the callopers. (b) Find the length of the line. |
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Answer» (a) Least Count(LC) `= ("Smallest division on main scale")/("Number of divisions on vernier scale")` `=(1)/(10) mm=0.1mm=0.01cm` (b) `L= N + n(LC)=(10.2+3xx 0.01)cm=10.23cm`. |
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| 25. |
In the diagram shown in figure, find the magnitude and nature of zero error. . |
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Answer» Here, zero of vernier scale lies to the right of zero of main scale, it has positive zero error. Futher, `N =0, x=5,LC or VC=0.01 cm` Hence, zero error =`N+x xx VC` `=0+5xx0.01=0.05cm` Zero correction =`-0.05 cm` ` :.` Actual length will be 0.05 cm less than the measured length. |
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| 26. |
In electrical calorimeter experiment, voltage across the heater is `100.0 V` and current is `10.0 A`. Heater is switched on for t `= 700.0 s`. Room temperature is `theta_0 = 10.0^(@)C` and final temperature of calorimeter and unknown liquid is `theta_f = 73.0^(@) C`. Mass of empty calorimeter is `m_1 = 1.0kg` and combined mass of calorimeter and unknown liquid is `m_2 = 3.0 kg`. Find the specificheat capacity of the unknown liquid in proper significant figures. Specific heat of calorimerter = `3.0xx10^3 j//kg .^(@) C`. |
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Answer» Correct Answer - A::C::D Given, `V= 100.0 V, i=10.0A, t=700.0 s,theta_0=10.0^(@) C,theta _f=73.0^(@) C,` `m_1=1.0kg` and `m_(2)=3.0kg` Substituting the values in the experssion. `S_(l)=(1/(m_(2)-m_(1)))[(Vit)/(theta_(f)-theta_(0))-m_(1)S_(c)]` we have, `S_(l)=1/(3.0-1.0)[((100.0)(10.0)(700.0))/(73.0-10.0)-(1.0)(3.0xx10^(3))]` `= 4.1xx 10^(3) j//kg .^(@) C` (According to the rules of significant figures). |
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| 27. |
The diagram below shows part of the main scale and vernier scale of a vernier callipers, which is used to measure the diameter of a metal ball. Find the least count and the radius of the ball. . |
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Answer» Correct Answer - A::B::C Least count ` = ("Value of one main scale division")/("Number of divisions on vernier scale")` = `(1 mm)/(10) = (0.1 cm)/(10)` = `0.01 cm` Reading (diameter) = MS reading + (coinciding VS reading xx Least Count) = `4.3 cm + (7 xx 0.01)` = `4.3 + 0.07 = 4.37 cm` Diameter = `4.37 cm` `:.` Radius =`(4.37)/2 = 2.185 cm` = `2.18 cm` To the required number of significant figures. |
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| 28. |
Screw gauge is said to have a negative error.A. when circular scale zero coincides with base line of main scale,B. when circular scale zero is above the base line of main scale.C. when circular scale zero is below the base line of main scale.D. None of the above. |
| Answer» Correct Answer - B | |
| 29. |
In the previous question, find the maximum permissible error in resistivity and resistance. |
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Answer» Correct Answer - A::B::D `(Delta rho)/rho xx 100 = [2(Deltad)/d+ (Delta V)/V+(Deltal)/l+(Delta I)/I] xx 100` = `[2 xx ((0.01)/(2.00))+((0.1)/(100.0))+((0.1)/(31.4))+((0.1)/(10.0))]xx 100` = `2.41 %` `R = V/I` ` rArr (DeltaR)/R xx 100 =[((Delta V)/V)+((DeltaI)/I)] xx 100` = `[(0.1)/(100)+0.1/(10)] xx 100 =1.1 %` . |
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| 30. |
In an experiment, current measured is, `1 = 10.0 A`, potential difference measured is `V = 100.0 C`, length of the wire is `31.4 cm` and the diameter of the wire ` 2.00 mm` (all in correct significant figures). Find resistivity of the wire in correct significant figures. [Take `pi = 3.14`, exact]. (a) (b) (c) (d) . |
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Answer» Correct Answer - A::D `rho = (pid^2V)/(4lI) =((3.14)(2.00 xx 10^-3)(100.0))/((4)(31.4)(10.0)xx10^-2` `1.00 xx 10^-4 Omega - m` (to three significant figures). |
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| 31. |
A meter bridge is set- up as shown in figure, to determine an unknown resistance `X` using a standard ` 10 Omega` resistor. The galvanometer shows null point when tapping - key is at `52 cm` mark. The end -corrections are `1 m` and `2 cm` respectively for the ends `A` and `B`. The determined values of `X` is . (a) `10.2 Omeaga` (b) `10.6 Omega` (c) `10.8 Omega` (d) `11.1 Omega`. |
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Answer» Correct Answer - B Using the concept of balanced, Wheatstone bridge, we have, `P/Q = R/S` `:. X/((52 +1))=(10)/((48 + 2))` `:. X=(10 xx 53)/(50) =10.6 Omega` `:.` Correct option is (b). |
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| 32. |
`R_(1),R_(2),R_(3)` are different values `R, A,B` and `C` are the null points obtained corresponding to `R_(1),R_(2) and R_(3)` respectively. For which resistor, the value of `X` will be the most accurate and why? . |
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Answer» Correct Answer - A::B::C Slide wire bridge is most sensitive when the resistance of all the four arms of bridge is same. Hence, (B) is the most accurate answer. |
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| 33. |
`AB` is a wire of uniform resistance. The galvanometer `G` shows no deflection when the length `AC = 20 cm` and `CB = 80 cm`. The resistance `R` is equal to. .A. `80 Omega`B. `10 Omega`C. `20 Omega`D. `40 omega` |
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Answer» Correct Answer - C `R/(80)=(20)/(80)` `:. R = 20 Omega`. |
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| 34. |
In post office box experiment, if `Q/P=1/(10)`. In (R ) if ` 142 Omega` is used then we get deflection towards right and if ` R =143 Omega` , then deflection is towards left. What is the range of unknown resistance? |
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Answer» Correct Answer - A::B::C::D `P/Q = R/X` `rArr X =(Q/P) R =(1/(10))R` `R ` lies between `142 Omega` and `143 Omega` There, the unknown resistance `X` lies between `14.2 Omega` and `14.3 Omega`. |
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| 35. |
To locate null point, deflection battery key `K_(1)` is pressed before the galvanometer key `K_(2)`. Explain why? |
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Answer» If galvanometer key `K_(2)` is pressed firdt then just after closing the battery key `K_(1)` current suddenly increase. So, due to self induction ,a large back emf is generated in the galvanometer, which mwy damage the galvanometer. |
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| 36. |
If we use `100 Omega` and `200 Omega` in place of `R` and `X` we get null pont deflection, `l = 33cm`. If we intercharge the resistors, the null point length is found to be `67 cm` Find end corrections `prop and beta`. |
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Answer» Correct Answer - A::C `alpha=(R_(2)l_(1)-R_(1)l_(2))/(R_(1)-R_(2))=((200)(33)-(100)(67))/(100-200)=1 cm` `beta=(R_(1)l_(1)-R_(2)l_(2))/(R_(1)-R_(2))-100` `=((100)(33)-(200)(67))/(100-200)-100` `= 1 cm`. |
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| 37. |
Why should the amplitude be small foe a simple pendulum experiment? |
| Answer» Because `T =2pisqrt(L//g)` is based on the assumption that `sin theta ~= theta` which is true only for small amplitude. | |
| 38. |
Does the time period depend upon the mass, the sixe and the material of the bob. |
| Answer» No, the time period does not depend on any of the given three properties of the bob. | |
| 39. |
What result do you expect in above experiment if by mistake, ammeter is connected in parallel with voltmeter and resistance as shown in figure? . |
| Answer» As ammeter has very resistance, therefore most of the current will pass through the ammeter so reading of ammeter will be very large. | |