- Radius of wheel P, r_P = 30 cm. Circumference of wheel P, C_P = 2 Π r_P = 2 Π 30 = 60Π cm. - Radius of wheel Q, r_Q = 40 cm. Circumference of wheel Q, C_Q = 2 Π r_Q = 2 Π 40 = 80Π cm. - Let N_P be the number of revolutions of wheel P and N_Q be the number of revolutions of wheel Q while travelling the same distance D. - The distance travelled is given by the number of revolutions multiplied by the circumference. - D = N_P C_P = N_Q C_Q. - We are given that N_P = N_Q + 5000. - Substitute this into the distance equation: (N_Q + 5000) C_P = N_Q C_Q (N_Q + 5000) 60Π = N_Q 80Π - Divide both sides by Π: (N_Q + 5000) 60 = N_Q 80 60N_Q + 300000 = 80N_Q 300000 = 80N_Q - 60N_Q 300000 = 20N_Q N_Q = 300000 / 20 = 15000. - The number of revolutions for wheel Q is 15000. - The distance travelled D = N_Q C_Q = 15000 80Π cm = 1200000Π cm. - We need the speed in km/hour. Convert distance from cm to km: 1 km = 100000 cm. D = 1200000Π cm / (100000 cm/km) = 12Π km. - Bicycle Q travelled this distance in 45 minutes. Convert time from minutes to hours: 45 minutes = 45/60 hours = 3/4 hours. - Speed of Q = Distance / Time Speed_Q = (12Π km) / (3/4 hours) = 12Π (4/3) km/hour = 4Π 4 km/ hour = 16Π km/hour.
- 89 hours / T_true = 89 / 90. - T_true = 90 hours. - Now, calculate the true time that is 90 hours after Monday 5 a.m. - Monday 5 a.m. + 24 hours = Tuesday 5 a.m. - Monday 5 a.m. + 48 hours = Wednesday 5 a.m. - Monday 5 a.m. + 72 hours = Thursday 5 a.m. - Monday 5 a.m. + 90 hours = (Monday 5 a.m. + 72 hours) + 18 hours = Thursday 5 a.m. + 18 hours. - 18 hours after 5 a.m. is 5 a.m. + 12 hours (to 5 p.m.) + 6 hours = 5 p.m. + 6 hours = 11 p.m. - The true time is 11 p.m. on Thursday. 276. The wheels of bicycles P and Q have radii 30 cm and 40 cm, respectively. While travelling a certain distance, wheel of P required 5000 more revolutions than wheel of Q. If bicycle Q travelled this distance in 45 minutes, then how much was its speed, in km per hour?
Asked in GPSC Prelims CSP 1 2023 · Topic: Mechanics (Newton's Laws of Motion and their applications)
More Physics questions
- If a journey of 120 km is done by train and the rest by car, then it takes 8 hours for total 600 km journey. And if 200 km is done by train and the rest by car, then it takes 20 minutes more. Then what would be the ratio of the speed of the train to that of the car?
- $\frac{120}{v_T} + \frac{480}{v_C} = 8$. (Equation 1)Case 2: 200 km by train, remaining by car. Remaining distance $= 600 - 200 = 400$ km by car. Total time = 8 hours 20 minutes. 8 hours 20 minutes $= 8 + \frac{20}{60}$ hours $= 8 + \frac{1}{3}$ hours $= \frac{25}{3}$ hours. Time by train + Time by car = $\frac{25}{3}$. $ \frac{200}{v_T} + \frac{400}{v_C} = \frac{25} {3}$. (Equation 2)Let $x = \frac{1}{v_T}$ and $y = \frac{1}{v_C}$. The equations become: $120x + 480y = 8$ (Equation 1') $200x + 400y = \frac{25} {3}$ (Equation 2')Divide Equation 1' by 8: $15x + 60y = 1$. (Equation 1'')From Equation 1'', $15x = 1 - 60y$.Substitute this into Equation 2': $200 \left(\frac{1 - 60y}{15}\right) + 400y = \frac{25} {3}$. $\frac{40}{3} (1 - 60y) + 400y = \frac{25} {3}$. Multiply the entire equation by 3: $40(1 - 60y) + 1200y = 25$. $40 - 2400y + 1200y = 25$. $40 - 1200y = 25$. $1200y = 40 - 25 = 15$. $y = \frac{15}{1200} = \frac{1}{80}$. Since $y = \frac{1}{v_C}$, $v_C = 80$ km/h.Substitute the value of $y$ back into Equation 1'': $15x + 60\left(\frac{1}{80}\right) = 1$. $15x + \frac{60} {80} = 1$. $15x + \frac{3}{4} = 1$. $15x = 1 - \frac{3}{4} = \frac{1}{4}$. $x = \frac{1}{4 \times 15} = \frac{1}{60}$. Since $x = \frac{1}{v_T}$, $v_T = 60$ km/h.The ratio of the speed of the train to that of the car is $v_T : v_C = 60 : 80$. Divide both parts by 20: $60 : 80 = \frac{60}{20} : \frac{80}{20} = 3 : 4$. 278. A speedboat runs 6 km upstream in a river and comes back to the starting point in 33 minutes. The stream of the river is running at 2 km/hr. What is the speed of speedboat in still water?
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