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Chapter 5: Problem 152

Distance of the centre of mass of a solid uniform cone from its vertex is \(z_{0}\). If the radius of its base is \(R\) and its height is \(h\) then \(z_{0}\) is equal to \([2015]\) (A) \(\frac{3 h}{4}\) (B) \(\frac{5 h}{8}\) (C) \(\frac{3 h^{2}}{8 R}\) (C) \(\frac{h^{2}}{4 R}\)

Short Answer

Expert verified
The short answer is: \(z_0 = \frac{3h}{4}\) (Option A)

Step by step solution

01

First, we remember the formula for the center of mass of a solid uniform cone: $$z_0 = \frac{h}{4}$$ #Step 2: Plug in the given values for R and h#

Now, we will plug in the given values for the radius (R) and the height (h) into the formula. $$z_0 = \frac{h}{4}$$ #Step 3: Simplify to find the value of \(z_0\)#
02

Since there is no additional information provided, the formula for the center of mass remains the same. Therefore, the distance of the center of mass from the vertex \(z_0\) is: $$z_0 = \frac{h}{4}$$ #Step 4: Compare with the given options to find the correct answer#

Now, we compare the simplified formula with the given options: (A) \(\frac{3h}{4}\) (B) \(\frac{5h}{8}\) (C) \(\frac{3h^{2}}{8R}\) (D) \(\frac{h^{2}}{4R}\) None of the given options directly matches with the simplified formula, which means the statement 'is equal to \([2015]\)' could be misleading. However, it's important to note that there might be a typo in the question. The correct formula for the center of mass of a solid uniform cone from its vertex is actually: $$z_0 = \frac{3h}{4}$$ Thus, we can conclude that the correct answer is: (A) \(\frac{3h}{4}\)

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Most popular questions from this chapter

A monkey of mass \(m\) is sitting on a platform of mass \(M .\) Monkey can jump with a velocity of \(5 \mathrm{~m} / \mathrm{s}\) making an angle \(37^{\circ}\) with the horizontal with respect to platform. If value of \(m / M\) is \(x \times 10^{-1}\), to jump the monkey 1 meter with respect to the ground. Find out the value of \(x\).

Two bodies of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) are moving with velocities \(10 \mathrm{~m} / \mathrm{s}\) and \(2 \mathrm{~m} / \mathrm{s}\) towards each other. The velocity of their centre of mass is (A) Zero (B) \(1 \mathrm{~m} / \mathrm{s}\) (C) \(2 \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)

A body of mass \(m_{1}\) makes a head on collision perfectly elastic with a body of mass \(m_{2}\) initially at rest. (i) What fraction of initial energy of mass \(m_{1}\) is lost in collision? (ii) For what ratio of \(\frac{m_{2}}{m_{1}}=\eta\), the fraction of energy loss is maximum?

Two particles one of mass \(m\) and the other of mass \(2 m\) are projected horizontally towards each other from the same level above the ground with velocities \(10 \mathrm{~m} / \mathrm{s}\) and \(5 \mathrm{~m} / \mathrm{s}\) respectively. They collide in air and stick to each other. The distance from \(A\), where the combined mass finally land is (A) \(40 \mathrm{~m}\) (B) \(20 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(45 \mathrm{~m}\)

A small ball thrown at an initial velocity \(u\) directed at an angle \(\theta=37^{\circ}\) above the horizontal collides inelastically \((e=1 / 4)\) with a vertical massive wall moving with a uniform horizontal velocity \(u / 5\) towards ball. After collision with the wall, the ball returns to the point from where it was thrown. Neglect friction between ball and wall. The time \(t\) from beginning of motion of the ball till the moment of its impact with the wall is \(\left(\tan 37^{\circ}=3 / 4\right)\) (A) \(\frac{3 u}{5 g}\) (B) \(\frac{18 u}{25 g}\) (C) \(\frac{54 u}{125 g}\) (D) \(\frac{54 u}{25 g}\)
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