| 1. | \( \theta=\sin ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1/2}\) | 2. | \(\theta=\sin ^{-1}\left(\frac{2 {gT}^2}{\pi^2 {R}}\right)^{1 / 2}\) |
| 3. | \(\theta=\cos ^{-1}\left(\frac{{gT}^2}{\pi^2 {R}}\right)^{1 / 2}\) | 4. | \(\theta=\cos ^{-1}\left(\frac{\pi^2 {R}}{{gT}^2}\right)^{1 / 2}\) |
1. \(\dfrac{r_1}{r_1+r_2}\)
2. \(\dfrac{r_2}{r_1+r_3}\)
3. \(\dfrac{r_1}{r_2+r_3}\)
4. \(\dfrac{r_2}{r_2+r_3}\)
A point object is placed at a distance of \(60~\text{cm}\) from a convex lens of focal length \(30~\text{cm}\). If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of \(40~\text{cm}\) from it, the final image would be formed at a distance of:
| 1. | \(30~\text{cm}\) from the plane mirror, it would be a virtual image. |
| 2. | \(20~\text{cm}\) from the plane mirror, it would be a virtual image. |
| 3. | \(20~\text{cm}\) from the lens, it would be a real image. |
| 4. | \(30~\text{cm}\) from the lens, it would be a real image. |
A uniform rod of length \(200~ \text{cm}\) and mass \(500~ \text g\) is balanced on a wedge placed at \(40~ \text{cm}\) mark. A mass of \(2~\text{kg}\) is suspended from the rod at \(20~ \text{cm}\) and another unknown mass \(m\) is suspended from the rod at \(160~\text{cm}\) mark as shown in the figure. What would be the value of \(m\) such that the rod is in equilibrium?
(Take \(g=10~( \text {m/s}^2)\)
| 1. | \({\dfrac 1 6}~\text{kg}\) | 2. | \({\dfrac 1 {12}}~ \text{kg}\) |
| 3. | \({\dfrac 1 2}~ \text{kg}\) | 4. | \({\dfrac 1 3}~ \text{kg}\) |
In the product
\(\vec{F}=q\left ( \vec{v}\times \vec{B} \right )=q\vec{v}\times \left ( B\hat{i}+B\hat{j}+B_0\hat{k} \right ).\) For \(q=1 \) and \(\vec{v}=2\hat{i}+4\hat{j}+6\hat{k} \) and \(\vec{F}=4\hat{i}-20\hat{j}+12\hat{k},\) what will be the complete expression for \(\vec{B} \)?
1. \(8\hat{i}+8\hat{j}-6\hat{k}\)
2. \(6\hat{i}+6\hat{j}-8\hat{k}\)
3. \(-8\hat{i}-8\hat{j}-6\hat{k}\)
4. \(-6\hat{i}-6\hat{j}-8\hat{k}\)
Find the correct sequence of the bond enthalpy of the "C-X" bond among the following:
1. CH3-F < CH3-Cl > CH3-Br > CH3-I
2. CH3-Cl > CH3-F > CH3-Br > CH3-I
3. CH3-F < CH3-Cl < CH3-Br < CH3-I
4. CH3-F > CH3-Cl > CH3-Br > CH3-I
Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of:
| 1. | Lanthanoid contraction |
| 2. | Having similar chemical properties |
| 3. | Belonging to the same group |
| 4. | Diagonal relationship |
\(C_6H_{6(l)}+\frac{15}{2}O_{2(\text{g)}}\longrightarrow~6CO_{2(\text{g)}}+3H_2O_{(l)}\)
1. \(11.2~\text{litre}\)
2. \(22.4~\text{litre}\)
3. \(84~\text{litre}\)
4. \(168~\text{litre}\)
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