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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.
| 401. |
Convert into lower units:(i) 15 km (m, cm, mm)(ii) 12 kg (g, mg) |
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Answer» (i) 15 km (m, cm, mm) 15 km = 15 × 1000 m = 15000 m (ii) 12 kg (g, mg) 12 kg = 12 × 1000 g |
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| 402. |
Convert 85 l 720 ml into ml, l and kl? |
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Answer» 85 l 720 ml (i) 85 l 720 ml = 85 × 1000 + 720 ml = 85000 + 720 ml = 85,720 (ii) 85 l 720 ml = 85 l + 720 × 11000 l = 85 l + 0.720 l = 85.72 l. (iii) 85 l 720 ml = 85.72 l = 85.72 × 11000 kl = 0.08572 kl. |
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| 403. |
2 days = …….. hours(i) 38(ii) 48(iii) 28(iv) 40 |
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Answer» (ii) 48 2 days = 48 hours |
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| 404. |
Which is greater 0.007 g, 0.7 kg or 70 mg? |
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Answer» 0.007 g = 0.007 × 1000 mg = 7 mg 0.7 kg = 0.7 × 10,00,000 mg = 7,00,000 mg 70 mg = 70 mg ∴ 0.7 kg is greater. |
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| 405. |
The duration of electricity consumed by the farmer for his pumpset on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity. |
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Answer» Total duration of electricity consumed on both the days = 7 hours 20 min 35 sec + 3 hours 44 min 50 sec = (7 + 3) hours (20 + 44) min (35 + 50) sec = 10 hours 64 min 85 seconds = 11 hours 5 min 25 seconds |
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| 406. |
The departure and arrival timing of the Vaigai Superfast Express (No. 12635) from Chennai Egmore to Madurai Junction are given. Read the details and answer the following.(i) At what time does the Vaigai Express start from Chennai and arrive at Madurai?(ii) How many halts are between there Chennai and Madurai?(iii) How long does the train halt at the Villupuram junction?(iv) At what time does the train come to Sholavandan?(v) Find the journey time from Chennai Egmore to Madurai? |
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Answer» (i) It starts from Chennai at 13 : 40 hrs and arrive at Madurai at 21 : 20 hrs. (ii) There are 8 halts. (iii) Departure from Villupuram = 15 hours 55 minutes Arrival at Villupuram = 15 hours 50 minutes The train halt at Villupuram for = 05 minutes (iv) At 20 : 34 hours the train come to Sholavandan (v) Arrival time at Madurai = 20 hours 80 (20 + 60) minutes Departure time from Chennai Egmore = 13 hours 40 minutes Journey Time = 07 hours 40 minutes |
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| 407. |
A bus leaves for Kanchipuram from Chennai at 4.30 p.m. It takes 1 hr 25 min. to reach there. At what time will it reach at Kanchipuram? |
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Answer» Starting time from Chennai = 4 : 30 pm. Duration of travel = 01 hr : 25 min Arrival at Kanchipuram = 5 : 55 pm The bus will reach Kanchipuram at 5 : 55 pm. |
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| 408. |
The duration of a film show is 3 hrs 15 min. It starts at 6 : 30 p.m. When will it end? |
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Answer» The film starts at = 6 : 30 pm. Duration of the film show = 3 hr : 15 min The show end at = 9 : 45 p.m. |
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| 409. |
A clock gains 3 minutes every hour. If the clock is set correctly at 5 a.m, find the time shown by the clock at 7 p.m? |
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Answer» 5 a.m. = 5.00 hours 7 p.m = 19.00 hours Difference = 19.00 – 5.00 = 14.00 hours Time gain = 14 × 3 = 42 minutes Time shown by the clock = 7.42 p.m |
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| 410. |
A train arrive Chennai Central at 11 : 55 am. It reached Chennai 1 hr 25 min late. What is the scheduled arrival time of the train at Chennai? |
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Answer» The train arrived Chennai at 11 : 55 am = 11 : 55 hours The train is late by = 1 hour 25 min Scheduled Time = 10 : 30 hours The scheduled arrival time = 10.30 a.m |
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| 411. |
Mammalia travelled 4 hrs 45 min by bus and 4 hrs 45 min by train. Calculate the time she spent in travelling? |
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Answer» Time travelled by bus = 4 hrs 45 min Time travelled by train = 4 hrs 45 min Total time she travelled = 8 hrs 90 min = 8 hrs (60 + 30) min 8 hrs + 1 hr 30 min Total time she travelled = 9 hr 30 min |
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| 412. |
Thenmozhi’s height is 1.25 m now. She grows for 5 cm every year. What would be her height after 6 years? |
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Answer» Thenmozhi’s present height = 1.25 m Rate of growth per year = 5 cm Her growth in 6 years = 5 cm × 6 = 30 cm. After 6 years her height = 1.25 m + 30 cm = 1.25 × 100 + 30 cm = 125 + 30 cm = 155 cm. ∴ After 6 years Thenmozhi’s height will be 155 cm. |
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| 413. |
Malar reached her school at 7 : 30 a.m. and left for home at 12 : 45 p.m. How long did she stay in school? |
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Answer» Malar left for home at 12 : 45 p.m. = 12 : 45 hrs She reached school at = 7 : 30 hrs She stayed at school for = 5 : 15 hrs |
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| 414. |
Priya bought 22 1/2 kg of onion/ Krishna bought 18 3/4 kg of onion and sethu bought 9 kg 250 g of onion, what is the total weight of onion did they buy? |
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Answer» Total weight of onion bought = 22 1/2 + 18 3/4 + 9 1/4 kg = 22 kg 500 g + 18 kg 750 g + 9 kg 250 g = 49 kg 1500 g = 50 kg 500 g |
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| 415. |
Sneha needs 1 m 20 cm of cloth for her blouse. If she bought 20 such cloths find the total cloth she bought in cm? |
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Answer» For one blouse the cloth needed = 1 m 20 cm Cloth needed for 20 blouse = 1 m 20 cm × 20 = 20 m 400 cm = 20 × 100 + 400 cm = 2000 + 400 cm = 2400 cm. Total cloth bought = 2400 cm. |
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| 416. |
Which among the following is (are) the units(s) of pressure?A. N `m^(-2)`B. gwt `cm^(-2)`C. kgwt `m^(-2)`D. All of the above |
| Answer» Correct Answer - D | |
| 417. |
A lorry full of fertiliser is bought to the garden having 120 coconut trees. If the lorry contain 1920 kg, then find the measure of fertiliser each coconut tree can have? |
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Answer» Total measure of fertiliser = 1920 kg. Number of coconut tress = 120 For one tree = 1920/120 kg = 16 kg. One tree can have 16 kg of fertiliser |
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| 418. |
Which of the following system of units is not based on units of mass, length and time alone?(A) S.I. (B) M.K.S (C) F.P.S (D) C.G.S |
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Answer» Answer is (A) S.I. Because in S.I. system, there are seven fundamental quantities. |
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| 419. |
The equation of state for a real gas having pressure P, volume V and temperature T is expressed as `(P+(a)/(V^(2)))(V-b)=RT` where a, b and R are constant. What are the dimensional formula of `(a)/(b)`?A. `[M^(2)L^(2)T^(-2)]`B. `[M^(1)L^(1)T^(-2)]`C. `[M^(1)L^(2)T^(-2)]`D. `[M^(2)L^(3)T^(-1)]` |
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Answer» Correct Answer - c In a dimensional equation, every term has the same dimensions as only quantities having the same dimensions can be added or subtracted. `because` In the term (v - b), b has the dimensions of V or `L^(3)`. Similarly dimensionally `P=(a)/(V^(2))` `therefore a=PV^(2)` `therefore [a]=[("Force")/("Area")xx("Area"xxL)^(2)]` `=[Fxx"Area"xxL^(2)]` `therefore [a]=[L^(1)M^(1)T^(-2)xxL^(2)xxL^(2)]=[L^(5)M^(1)T^(-2)]` `therefore [(a)/(b)]=[(L^(5)M^(1)T^(-2))/(L^(3))]=[M^(1)L^(2)T^(-2)]` |
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| 420. |
If dimensions of A and B are different, then which of the following operaion is valid ?A. `(A)/(B)`B. `e^(-A//B)`C. A-BD. A+B |
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Answer» Correct Answer - A `(A)/(B)` operation is valid for A and B having different dimension. |
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| 421. |
Fill in the blanks.(i) 250 ml + 1/2 ml = _____ l.(ii) 150 kg 200 g + 55 kg 750 g = ____ kg ____ g.(iii) 20 l – 1 l 500 ml = ____ l ___ ml(iv) 450 ml × 5 = ____ l ____ ml.(v) 50 Kg ÷ 100 g = ______ |
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Answer» (i) 3/4 l (ii) 205 kg 950 g (iii) 18 l 500 ml (iv) 2l 250 ml (v) 500 |
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| 422. |
Every day 150 l of water is sprayed in the garden. Water sprayed in a week is ____(a) 700 l(b) 1000 l(c) 950 l(d) 1050 l |
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Answer» (d) 1050 l Every day 150 l of water is sprayed in the garden. Water sprayed in a week is 1050 l. |
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| 423. |
If u1 and u2 are the units selected in two systems of measurement and n1 and n2 are their numerical values, then (A) n1u1 = n2u2 (B) n1u1 + n2u2 = 0 (C) n1n2 = u1u2 (D) (n1+ u1) = (n2 + u2) |
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Answer» Answer is (A) n1u1 = n2u2 Physical quantity (M) = Numerical value (n) x Unit (u) If physical quantity remains constant then n ∝1/u ∴ n1u1 = n2u2. |
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| 424. |
What is the order of magnitude of 1 light year expressed in metre?A. `10^(14)`B. `10^(15)`C. `10^(16)`D. `10^(17)` |
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Answer» Correct Answer - c 1 Light Year = 1 Year `xx` Velocity of light `=365xx24xx60xx60xx3xx10^(8)` metre `=9.461xx10^(15)m` `therefore` The order of magnitude is `10^(16)`m. |
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| 425. |
In a JRC one day camp, 150 gm of rice and 15 ml oil are needed for a student. If there are 40 students to attend the camp how much of rice and oil are needed? |
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Answer» Rice needed for a student = 150 gm Rice needed for 40 students = 40 × 150 gm = 6000 gm = 6 kg Oil needed for a student = 15 ml Oil needed for 40 students = 40 × 15 ml = 600 ml |
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| 426. |
The magnitude of any physical quantity can be expressed as A x 10n where n is a number called order of magnitude and A is (A) 0.1 ≤ A < 1 (B) 0.5 ≤ A < 5 (C) 5 ≤ A < 9 (D) 1 ≤ A > 9 |
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Answer» Answer is (B) 0.5 ≤ A < 5 |
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| 427. |
One second is `((1)/(k))` th part of a mean solar day, where k = __________. |
| Answer» Correct Answer - 86400 | |
| 428. |
If the volume of 1 kg of oil, at a given temperature, is `1000 cm^(3)` , then the volume 1000 kg oil at the same temperature is _____. |
| Answer» Correct Answer - `10^(6) cm^(3)` | |
| 429. |
Assertion (A): A physical quantity which can be described with both magnitude and direction is called as a vector quantity. Reason (R): All vector quantities are derived quantities.A. A and R are correct and R is the correct explanation for AB. A and R are correct but R is not the correct explanation for AC. A is correct but R is incorrect.D. Both A and R are incorrect. |
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Answer» Correct Answer - C All vector quantities need to be derived quantities. For example, displacement is a vector quantity but it is not a derived quantity. The statement give as a person as a reason is correct. But the assertion is correct. |
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| 430. |
On droping 10 lead shots into a measuring jar containing a liquid, the level of the liquid in the jar increased from `40 cm^(3)` to `100 cm^(3)`. Then, the volume of each lead shot is __________. |
| Answer» Correct Answer - `6 cm^(3)` | |
| 431. |
If 10 cm is taken as unit of length, then the height of a person 1.78 m tall in the new system of units is _______ units. |
| Answer» Correct Answer - 17.8 | |
| 432. |
Which of the following equations is are true with reference to a vernier calipers?A. Least count = M.S.D. - 1 V.S.D.B. 1 M.S.D. = (L.C.) `xx` number of divisions on the Vernier scaleC. If the least is 0.1 mm, then M.S.D.= `N/10`mmD. All of the above |
| Answer» Correct Answer - D | |
| 433. |
While finding the internal diameter of a pipe using a vernier calipers, the M.S.R. = 18 mm, the vernier scale coinciding division is 3. If the least count of the instrument is 0.1 mm, then the internal radius is _________ mm. |
| Answer» Correct Answer - 9.15 | |
| 434. |
The following information is given in respect of a vernier calipers. 1 main scale division = 0.3 cm 30 V.S.D. = 29 M.S.D. The least count of this vernier calipers is ______________.A. 0.01 mmB. 0.1 cmC. 0.03 cmD. 0.1 mm |
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Answer» Correct Answer - D 1 MSD = 0.3 cm Since, 30 VSD = 29 MSD (i.e., N VSD = (N-1) MSD) Least count = `(1MSD)/(30)=(0.3 cm)/(30)` = 0.01 cm = 0.1 mm |
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| 435. |
While measuring the diameter of a sphere using a vernier calipers the main scale reading was found to be p while the vernier coinciding division was q. If the least count of the instrument is r, then the radius of the sphere is given by the expression ________.A. p+q+rB. `(qr+p)/(2)`C. p+qrD. `(pr+q)/(2)` |
| Answer» Correct Answer - B | |
| 436. |
The length of one main scale division of a given vernier calipers is 1 cm . When the jaws are in contact the last division of the vernier scale coincides with 99 th mark of the main scale. Then the least count of the calpers is `"________"`A. 0.01 mmB. 0.01 cmC. 0.1 cmD. 0.1 mm |
| Answer» Correct Answer - B | |
| 437. |
If p divisions on the venier scale of vernier calipers correspond to q main scale divisions `(p gt q)`, derive an expression ofr the least count of the instrument in terms of p and q. If 1 M.S.D. = k mm, express the least count in cm. |
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Answer» (i) Least count = 1 M.S.D. - 1 V.S.D. (ii) L.c.`= {(p-q)/(p)}M.S.D. = (k(p-q))/(10p)cm` |
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| 438. |
The internal jaws of a vernier calipers were held tightly inside a hollow tube when the zero of the vernier scale showed a reading of 1.9 cm and the 18th division on the vernier scale. The thickness of the wall of the tube is 1 mm. If M.S. D. = 1 mm and the number of division on the vernier scale is 20, find the outside diameter of the hallow tube. |
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Answer» (i) Outside diameter = Inside diameter + `(2 xx "thickness")` (ii) 21.9 mm |
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| 439. |
When the jaws of a vernier callipers are closed the 0th division of its vernier scale is to the right of the zero of the main scale and the V.S.D is 6 .Find the correction to be made to the observed measurement (take its least count as 0.1 mm) |
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Answer» The vernier coincidence ,n=6 Zero error =V.S.D `xx`L.C `=6xx 0.1 mm=0.6` and the error is positive .Thus the correction =-0.6 mm |
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| 440. |
A metabollic rod of unknown radius was held in between the lower jaws of a vernier callipers. The main scaleis divided into millimetres and the venier scale has 50 divisions. 9he vernier coincinding division is 39. If the zeroth division of the vernier scale lies between 6.3 cm and 6.4 cm of the main scale, find the radius. |
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Answer» Number of V.S.D., N = 50 V.C.D. = 39 M.S.R. = 6.3 cm = 63 mm Diameter of the metallic rod = 2R = `M.S.R + V.C.D xx L.C.` rArr `R=(M.S.R.+V.C.Dxx L.C.)/(2)` L.C. of the vernier callipers, L.C. = `(1M.S.D)/(N) =(1mm)/(50)=(0.1)/(5)=0.02 mm` Diameter =` 65 mm = 39 xx 0.02 mm` = (63 + 0.78)mm D = 63.78 mm (or) D 6.378 cm `therefore "Radius of the given rod"`, `R=(D)/(2)=(6.378)/(2)=3.189 cm = 31.89mm` |
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| 441. |
When an unknown physical quantity X is divided by the force, acting on the body, we get the velocity of the body. What is X ?A. WorkB. Kinetic energyC. PowerD. Linear momentum |
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Answer» Correct Answer - c `(X)/("Force")` = velocity `therefore X = Fv` But Power = Fv `therefore` X = Power |
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| 442. |
Match the following.PrefixSymbola. Centii. kb. Nanoii. cc. miliiii. nd. kiloiv. m |
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Answer» a. ii b. iii c. iv d. i |
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| 443. |
The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with vernier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will beA. `1.5%`B. `2.55`C. `3.5%`D. `4%` |
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Answer» Correct Answer - B Volume of cylinder `V=pir^(2)L,r=((D)/(2))` `:.((DeltaV)/(V))xx100=2((DeltaD)/(D))xx100+((DeltaL)/(L))xx100` `=2((0.01)/(4.0))xx100+((0.1)/(5))xx100=2.5%` |
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| 444. |
The number of significant figures in 0.06900 isA. 5B. 4C. 2D. 3 |
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Answer» Correct Answer - B `0.06900`=four significant figure. Figuress since zeroes on left side before a non-zero number are not significant figures. |
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| 445. |
A cuboid has volume `V=lxx2lxx3l`, where l is the length of one side. If the relative percentage error in the measurment of l is 1%, then the relative percentage error in measurement of V isA. 0.18B. 0.06C. 0.03D. 0.01 |
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Answer» Correct Answer - C Volume of cuboid, `V=lxx2lxx3l=6l^(3):.(DeltaV)/(V)xx100=3((Deltal)/(l))=3% [:.(Deltal)/(l)xx100=1%]` |
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| 446. |
The length of a rod as measured in an experiment was found to be 2.48m, 2.46 m, 2.49 m, 2.50 m and 2.48m. Find the average length, absolute arror in each observation and the percentage error.A. `(2.48pm0.01)m, 0.40%`B. `(3.48pm0.02)m,0.45%`C. `(2.48pm0.01)m,0.45%`D. `(3.48pm0.02)m,045%` |
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Answer» Correct Answer - A Average lrngth =Arithmetic mean of the measured values `x_(mean)=(2.48+2.46+2.49+2.49+2.46)/(5)=(12.38)/(5)=2.476m` `:.` True value, `X_(mean)=2.48m` Absolute error in various measurement , `|Deltax_(1)|=|x_(1)-x_("mean")|=2.48-2.48=0.00m` `|Deltax_(2)|=|2.46-2.48|=0.02m` `|Deltax_(3)|=|2.49-2.48|=0.01m` `|Delta x_(4)|=|2.49-2.48|=0.01m` `|Deltax_(5)|=|2.46-2.48|=+0.02m` Mean absolute error `=(|Deltax_(1)|+|Delta x_(2)|+|Deltax_(3)|+...+|Deltax_(5)|)/(5)` `=((0.00+0.02+0.01+0.01+0.02))/(5)=(0.06)/(5)` `Deltax_("mean")=0.01m` Thus, `x=2.48pm0.01m` Precentage error,`deltax=(Deltax_("mean"))/(x)xx100=(0.01)/(2.48)xx100=0.40%` |
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| 447. |
Calculate focal length of a spherical mirror from the following observations : object distance, `u = (50.1 +- 0.5)cm` and image distance , `upsilon = (20.1 +- 0.2)cm.`A. `(0.04pm14.3)`B. `(14.3pm0.4)`C. `(14.3pm0.04)`D. None of these |
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Answer» Correct Answer - B Formula forfocal length of a spherical mirror `(1)/f=(1)/(v)+(1)/(u)` `or f=(uv)/(u+v)` `=((50.1)(20.1))/((50.1+20.1))=14.3cm` Also `(Deltaf)/(f)=pm[(Deltau)/(u)+(Deltav)/(v)+(Deltau+Deltav)/(u+v)]` `=pm[(0.5)/(50.1)+(0.2)/(20.1)+(0.5+0.2)/(50.1+20.1)]` `=[0.00998+0.00995+0.00997]` `=pm(0.0299)` `:.Deltaf=0.0299xx14.3=0.428=0.4cm` `:.f=(14.3pm0.4)cm` `(DeltaT)/Txx100=pm((1)/(2)xx(Deltal)/(l)xx100+(1)/(2)xx(Deltag)/(g)xx100)` `=pm((1)/(2)xx1+(1)/(2)xx2)=pm1.5%` |
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| 448. |
Two resistances R1 = 50 ± 2 ohm and R2 = 60 ± 3 ohm are connected in series, the equivalent resistance of the series combination is (A) (110 ± 4) ohm (B) (110 ± 2) ohm (C) (110 ± 5) ohm (D) (110 ± 6) ohm |
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Answer» Answer is (C) (110 ± 5) ohm |
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| 449. |
The volumes of two bodies are measured to be `V_1 = (10.2+-0.02) cm^3 and V_2 = (6.4 +- 0.01)cm^3`. Calculate sum and difference in volumes with error limits.A. `(16.6pm0.03)(3.8pm0.03)`B. `(0.03pm3.8)(0.03pm16.6)`C. `(16.6pm3.8)(16.6pm0.03)`D. `(3.8pm0.05)(16.6pm0.03)` |
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Answer» Correct Answer - A Given, `V_(1)=(10.2pm0.02)cm^(3)` and `V_(2)=(6.4pm0.01)cm^(3)` `DeltaV=pm(DeltaV_(1)+DeltaV_(2))` `=pm(0.02+0.01)cm^(3)=pm0.03cm^(3)` `V_(1)+V_(2)=(10.2+6.4)cm^(3)=16.6cm^(3)` and `V_(1)-V_(2)=(10.2-6.4)cm^(3)=3.8cm^(3)` Hence, sum of volumes `=(16.6pm0.03)cm^(3)` and difference of volumes `=(3.8pm0.03)cm^(3)` |
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| 450. |
A simple pendulum can be used to determine acceleration due to gravity at a given place. |
| Answer» Correct Answer - True | |