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Class 11 NEB model question solution 2077 | Physics | Complete explanation and notes

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Model Question Solution Group 'A' 1. A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should this result be recorded in a table of results? (a) 0.2000 m (b) 0.200 m (c) 0.20 m (d) 0.2 m Explanation: From the question, it is obvious that the least count of the metre scale is 1 mm as it measures upto the millimeter. Since, 1 mm = 0.001 m . So, result should have three significant figures. Thus, \[\begin{align*} 20 cm &= \frac{20}{100}\\ \therefore 20 cm &=0.200 \hspace{0.01cm} m\\ \end{align*}\] 2. Forces are applied to a rigid body. The forces all act in the same plane. In which diagram is the body in equilibrium? (b) Explanation: In this case, the downward force has balanced the upward force and hence the body is in equilibrium. 3. An athlete makes a long jump a
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Class 12 NEB Physics Question Solution | 2079 (2022) Question Paper Solution | Physics

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NEB Grade XII 2079 (2022) Physics Question Solution New Course Group 'A' Rewrite the correct options of each question in your answer sheet. If L represents momentum, I represents moment of inertia, then \(\frac{L^2}{2I}\) represents Rotational kinetic energy Torque Power Potential energy Solution: Option (a) Explanation: Since, the rotational kinetic energy is commonly expressed as, \[\text{K.E.}=\frac{1}{2}mv^2\] Also, \[L=I\omega\] Then, \[\begin{align*} \text{K.E.}&=\frac{1}{2} I \times \frac{L^2}{I^2}\\ &=\frac{L^2}{2I}\\ \end{align*}\] Starting from mean position, a particle in simple harmonic motion takes time \(T_1\) and \(T_2\) to cover first half and next half displacement in moving from mean to extreme position, then
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Day pdf | NEB Exam Preparation.

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Class 12 Model Question Solution | Physics | 2078/2079 Model Question Solution

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Model Question Solution Group 'A' Which of the following is a correct formula for calculating radius of gyration of a rotating object? Solution: Option (a) k\(^2=\frac{I}{m}\) Explanation \[I=MK^2\] \[\therefore K^2=\frac{I}{M}\] A horizontal stream of air is blown under one of the pan balance as shown in the figure. What will be the effect on the pan? Solution: Option (b) goes down Explanation Bernoulli's theorem for the stream line flow of an ideal liquid (incompressible and non viscous) gives, \[\begin{align} \frac{P}{\rho} + gh + \frac{v^2}{2}&=\text{constant}\\ \frac{P}{\rho} + \frac{v^2}{2}& =\text{constant} \hspace{0.1cm} \because \text{two pans are at same height}\\ \end{align}\] When v goes up, P goes down. For the above expression to be constant. So, pressure above the pan is high and that below is low. D
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