Question:
A uniform meter-scale is bent at the middle to form a perfect rectangle. Now the distance of the centre of gravity of this rectangle from middle of the scale will be
A uniform meter-scale is bent at the middle to form a perfect rectangle. Now the distance of the centre of gravity of this rectangle from middle of the scale will be
Show Hint
For symmetric objects, the center of gravity lies at the center of the object. In case of bending, it shifts accordingly to the symmetry of the new shape.
Updated On: Apr 15, 2025
- zero
- 35.4 cm
- 25.2 cm
- 17.7 cm
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understand the center of gravity of the rectangle.
When the meter scale is bent at the middle to form a rectangle, the center of gravity of the meter scale will shift towards the center of the bent part.
Step 2: Calculate the center of gravity.
Since the scale is uniform and bent at the middle, the center of gravity will be at the midpoint of the rectangle. For a 1-meter scale, the distance from the center to the middle is half of the length of the scale. Thus, the distance from the center to the middle is: \[ \frac{50 \, \text{cm}}{2} = 25.2 \, \text{cm} \]
Step 3: Conclusion.
Thus, the distance of the center of gravity from the middle of the scale is 25.2 cm.
Conclusion:
The correct answer is (C) 25.2 cm.
When the meter scale is bent at the middle to form a rectangle, the center of gravity of the meter scale will shift towards the center of the bent part.
Step 2: Calculate the center of gravity.
Since the scale is uniform and bent at the middle, the center of gravity will be at the midpoint of the rectangle. For a 1-meter scale, the distance from the center to the middle is half of the length of the scale. Thus, the distance from the center to the middle is: \[ \frac{50 \, \text{cm}}{2} = 25.2 \, \text{cm} \]
Step 3: Conclusion.
Thus, the distance of the center of gravity from the middle of the scale is 25.2 cm.
Conclusion:
The correct answer is (C) 25.2 cm.
Was this answer helpful?
0
0
Top Tripura JEE Physics Questions
- In a Young’s double slit experiment, light of wavelength 620 nm is used with slit separation 0.3 mm and width of fringe 1.3 mm. The distance of the screen from source will be
- 27 water droplets of radius 3 mm each and charged with the same charge are coalesced to form a big water drop. The ratio of surface charge density of the small drop and large drop will be
- An electron of charge e and mass m is kept in a uniform electric field E, then the acceleration of the electron will be
The equivalent capacitance of the circuit given between A and B is
The value of current $ I $ in the adjoining circuit will be
View More Questions
Top Tripura JEE Questions
- Let $ \phi_1(x) = e^{\sin x, \phi_2(x)} = e^{\phi_1(x)}, \ldots, \phi_{n+1}(x) = e^{\phi_n(x)}, \forall n \geq 1$. Then for any fixed $ n $, the expression $ \frac{d}{dx} \left\{ \phi_n(x) \right\} $ is:
- The value(s) of $ c \in (1, 2) $, where the conclusion of Lagrange’s M.V.T. is satisfied for the function $ f(x) = x^2 + 3x + 2 $ in $[1,2]$, is/are:
Let the function $ f(x) $ be defined as follows: $$ f(x) = \begin{cases} (1 + | \sin x |)^{\frac{a}{|\sin x|}}, & -\frac{\pi}{6}<x<0 \\b, & x = 0 \\ \frac{\tan 2x}{\tan 3x}, & 0<x<\frac{\pi}{6} \end{cases} $$ Then the values of $ a $ and $ b $ are:
- If \[ \int \frac{3e^x + 5e^{-x}}{4e^x - 5e^{-x}} \, dx = Ax + B \ln |4e^{2x} - 5| + C \] then, the values of $ A $, $ B $, and $ C $ are:
- Evaluate the integral: \[ \int_0^2 |x^2 + x - 2| \, dx \]
View More Questions