Page 1 Section I Single Correct Option 1, K mv = 1 2 2 \ D D D K K m m v v æ è ç ö ø ÷ = + max 2 \ Maximum error = 2% + 2 3 ( %) In the estimate of, kinetic energy ( ) % K = 8 Option (c) is correct. 2. d m V m l = = 3 \ D ´ æ è ç ö ø ÷ = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ d d m m l l 100 100 3 100 max D D = + 4 3 3 % ( %) = 13% Option (d) is correct. 3. p F A F L = = 2 \ Permissible error in pressure ( ) p = + 4 2 2 % ( %) = 8% Option (a) is correct. 4. p V p V 1 1 2 2 = Þ p p V V 2 1 1 2 = = - p V V V 1 1 1 1 10% of = p 1 90% p p 2 1 9 = \ Percentage increase in pressure = 100 9 = 111 . Option (a) is correct. 5. K p m = 2 2 \ Error in the measurement of kinetic energy (K) = ´ 2 100% = 200% Option (d) is correct. 6. 3400 = ´ 3 400 10 3 . \ Number of significant figures = 2 Option (d) is correct. 7. A = ´ 3 124 3 002 . . m m = 9.378 / 248 m 2 = 9.378 m 2 Option (a) is correct. 8. g GM R = 2 = constant R 2 K I = 1 2 2 w = æ è ç ö ø ÷ 1 2 2 5 2 MR w = constant ´ R 2 Basic Mathematics & Measurements 1 Page 2 Section I Single Correct Option 1, K mv = 1 2 2 \ D D D K K m m v v æ è ç ö ø ÷ = + max 2 \ Maximum error = 2% + 2 3 ( %) In the estimate of, kinetic energy ( ) % K = 8 Option (c) is correct. 2. d m V m l = = 3 \ D ´ æ è ç ö ø ÷ = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ d d m m l l 100 100 3 100 max D D = + 4 3 3 % ( %) = 13% Option (d) is correct. 3. p F A F L = = 2 \ Permissible error in pressure ( ) p = + 4 2 2 % ( %) = 8% Option (a) is correct. 4. p V p V 1 1 2 2 = Þ p p V V 2 1 1 2 = = - p V V V 1 1 1 1 10% of = p 1 90% p p 2 1 9 = \ Percentage increase in pressure = 100 9 = 111 . Option (a) is correct. 5. K p m = 2 2 \ Error in the measurement of kinetic energy (K) = ´ 2 100% = 200% Option (d) is correct. 6. 3400 = ´ 3 400 10 3 . \ Number of significant figures = 2 Option (d) is correct. 7. A = ´ 3 124 3 002 . . m m = 9.378 / 248 m 2 = 9.378 m 2 Option (a) is correct. 8. g GM R = 2 = constant R 2 K I = 1 2 2 w = æ è ç ö ø ÷ 1 2 2 5 2 MR w = constant ´ R 2 Basic Mathematics & Measurements 1 \ Decrease in R (radius) by 2% world increase g by 4% and decrease K (rotational kinetic energy) by 4%. Option (b) is correct. 9. Heat (H) = 2 2 Rt \ Maximum error in measuring heat (H) = + + 2 2 1 1 ( %) % % = 6% Option (b) is correct. 10. V lbt = = ´ ´ 12 6 2 45 . = 176 4 . = ´ 1764 10 2 3 . cm = ´ 2 10 2 2 cm Option (b) is correct. 11. I P r = 4 2 p i.e., Ir 2 = constant i.e., if r is increased by 2% the intensity will decrease by 4%. Option (d) is correct. 12. Option (b) is correct. 13. V r = 4 3 3 p D D V V r r = 3 = 3 1 ( %) = 3% Option (c) is correct. 14. a a 3 2 6 = (given) \ a = 6 Þ V = = 6 216 3 3 m Option (b) is correct. 15. g l T = 4 2 2 p \ D D D g g l l T T ´ æ è ç ö ø ÷ = ´ + ´ æ è ç ö ø ÷ 100 100 2 100 max = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 1 100 100 2 0 15 2003 10 mm m . = + 0 1 0 1 . % . % = 0.2 % Option (a) is correct. 16. Q I e tI V = - e a b ( ) D 0 (given) We know that Q It = \ t e tI V = - e a b ( ) D 0 Þ [ ] [ ] a = t and [ ] ( ) b = e é ë ê ù û ú tI V D 0 Þ b a é ë ê ù û ú = e é ë ê ù û ú = é ë ê ù û ú e é ë ê ù û ú tI V t I V ( ) ( ) D D 0 0 1 = e é ë ê ù û ú 1 1 0 [Resistance] or b a é ë ê ù û ú = 1 1 [ML T A [M L T A ] 2 â€“3 â€“2 â€“1 â€“3 4 2 ] = = 1 [L T] â€“1 [velocity] = e 1 0 1 2 [ ] / m 0 Option (a) is correct. 17. a pq r s = 2 3 Da a p p q q ´ æ è ç ö ø ÷ = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 100 1 2 100 1 2 100 max D D + ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 2 100 3 100 D D r r s s = + 1 2 1 1 2 3 ( %) ( %) + + 2 3 3 ( %) ( %) 0.5 0.3 = + 0.5% 1.5% + + 1 1 % % = 4% Option (c) is correct. 18. Least count of main scale = = 2 4 mm 0.5 mm Least count = least count of main scale 50 = 0.1 mm Zero error = - ´ 30 0.01 mm = -0.3 mm (â€“ive sign, zero of circular scale is lying above observed reading of plate thick) = 2 MSR + 20 CSR Basic Mathematics & Measurements | 3 Page 3 Section I Single Correct Option 1, K mv = 1 2 2 \ D D D K K m m v v æ è ç ö ø ÷ = + max 2 \ Maximum error = 2% + 2 3 ( %) In the estimate of, kinetic energy ( ) % K = 8 Option (c) is correct. 2. d m V m l = = 3 \ D ´ æ è ç ö ø ÷ = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ d d m m l l 100 100 3 100 max D D = + 4 3 3 % ( %) = 13% Option (d) is correct. 3. p F A F L = = 2 \ Permissible error in pressure ( ) p = + 4 2 2 % ( %) = 8% Option (a) is correct. 4. p V p V 1 1 2 2 = Þ p p V V 2 1 1 2 = = - p V V V 1 1 1 1 10% of = p 1 90% p p 2 1 9 = \ Percentage increase in pressure = 100 9 = 111 . Option (a) is correct. 5. K p m = 2 2 \ Error in the measurement of kinetic energy (K) = ´ 2 100% = 200% Option (d) is correct. 6. 3400 = ´ 3 400 10 3 . \ Number of significant figures = 2 Option (d) is correct. 7. A = ´ 3 124 3 002 . . m m = 9.378 / 248 m 2 = 9.378 m 2 Option (a) is correct. 8. g GM R = 2 = constant R 2 K I = 1 2 2 w = æ è ç ö ø ÷ 1 2 2 5 2 MR w = constant ´ R 2 Basic Mathematics & Measurements 1 \ Decrease in R (radius) by 2% world increase g by 4% and decrease K (rotational kinetic energy) by 4%. Option (b) is correct. 9. Heat (H) = 2 2 Rt \ Maximum error in measuring heat (H) = + + 2 2 1 1 ( %) % % = 6% Option (b) is correct. 10. V lbt = = ´ ´ 12 6 2 45 . = 176 4 . = ´ 1764 10 2 3 . cm = ´ 2 10 2 2 cm Option (b) is correct. 11. I P r = 4 2 p i.e., Ir 2 = constant i.e., if r is increased by 2% the intensity will decrease by 4%. Option (d) is correct. 12. Option (b) is correct. 13. V r = 4 3 3 p D D V V r r = 3 = 3 1 ( %) = 3% Option (c) is correct. 14. a a 3 2 6 = (given) \ a = 6 Þ V = = 6 216 3 3 m Option (b) is correct. 15. g l T = 4 2 2 p \ D D D g g l l T T ´ æ è ç ö ø ÷ = ´ + ´ æ è ç ö ø ÷ 100 100 2 100 max = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 1 100 100 2 0 15 2003 10 mm m . = + 0 1 0 1 . % . % = 0.2 % Option (a) is correct. 16. Q I e tI V = - e a b ( ) D 0 (given) We know that Q It = \ t e tI V = - e a b ( ) D 0 Þ [ ] [ ] a = t and [ ] ( ) b = e é ë ê ù û ú tI V D 0 Þ b a é ë ê ù û ú = e é ë ê ù û ú = é ë ê ù û ú e é ë ê ù û ú tI V t I V ( ) ( ) D D 0 0 1 = e é ë ê ù û ú 1 1 0 [Resistance] or b a é ë ê ù û ú = 1 1 [ML T A [M L T A ] 2 â€“3 â€“2 â€“1 â€“3 4 2 ] = = 1 [L T] â€“1 [velocity] = e 1 0 1 2 [ ] / m 0 Option (a) is correct. 17. a pq r s = 2 3 Da a p p q q ´ æ è ç ö ø ÷ = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 100 1 2 100 1 2 100 max D D + ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 2 100 3 100 D D r r s s = + 1 2 1 1 2 3 ( %) ( %) + + 2 3 3 ( %) ( %) 0.5 0.3 = + 0.5% 1.5% + + 1 1 % % = 4% Option (c) is correct. 18. Least count of main scale = = 2 4 mm 0.5 mm Least count = least count of main scale 50 = 0.1 mm Zero error = - ´ 30 0.01 mm = -0.3 mm (â€“ive sign, zero of circular scale is lying above observed reading of plate thick) = 2 MSR + 20 CSR Basic Mathematics & Measurements | 3 = ´ ( . 2 05 mm) + (20 × 0.01 mm) = 1 mm + 0.2 mm = 1.2 mm. Plate thickness (corrected reading) = observed reading - zero error = 1.2 mm + 0.3 mm = 1.5 mm Option (d) is correct. More than One Correct Options 1. Maximum percentage error in x : = + (% ) (% ) error in error in a b 2 +3 (% error in c) = 15% Assertion and Reason 1. Least count of screw gauge = Pitch Number of divisions of circular scale Less the value of pitch, less will be least count of screw gauge leading to len uncertainty that is more accuracy in the measurement. Thus, assertion is true. From the above relation we conclude that least count of screw gauge is inversely proportional to the number of divisions of circular scale. Thus reason is false. Option (c) is correct. Match the Columns 1. (a) F GM M r = 1 2 2 \ GM M Fr 1 2 2 = i e . ., [ ] [ GM M 1 2 = MLT ][L ] â€“2 2 = [ML T ] 3 â€“2 \ ( ) ( ) a q ® (b) 3 3 3 RT M pV n nM = = work \ 3 ML T M [L T ] 2 2 2 2 RT M é ë ê ù û ú = é ë ê ù û ú = - - \ (b) ® (r) (c) F q B F qB 2 2 2 2 = æ è ç ö ø ÷ = æ è ç ö ø ÷ qvB qB sinq 2 = v 2 2 sin q \ F q B 2 2 2 é ë ê ù û ú = [LT ] = [L T ] â€“1 2 2 â€“2 \ (c) ® ( ) r (d) g GM R e e = 2 GM R gR e e e = Þ GM R e e é ë ê ù û ú = [LT ][L] â€“2 = [L T ] 2 â€“2 \ (d) ® ( ) r 4 | Mechanics-1Read More

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