In a vernier callipers, \(20\) VSD coincides with \(16\) MSD (each division of length \(1\) mm). The least count of the vernier callipers is:
1. \(0.2\) cm
2. \(0.1\) cm
3. \(0.02\) cm
4. \(0.01\) cm
1. \(0.2\) cm
2. \(0.1\) cm
3. \(0.02\) cm
4. \(0.01\) cm
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Each side of a metallic cube of mass \(5.580\) kg is measured to be \(9.0\) cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as \(X \times10^3 ~\text{kg m}^{-3},\) where the value of \(X\) is:
1. \(7.654\)
2. \(7.7\)
3. \(7.65\)
4. \(7.6~\)
1. \(7.654\)
2. \(7.7\)
3. \(7.65\)
4. \(7.6~\)
Level 4: Below 35%
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A ray of monochromatic light is passing through an equilateral prism \((ABC)\) as shown in the figure. The refracted ray \((QR)\) is parallel to its base \((BC)\) and the angle of incidence \((i)\) is \(50^\circ\). Then the angle of deviation \((\delta)\) is:
1. \(45^\circ\)
2. \(55^\circ\)
3. \(35^\circ~\)
4. \(40^\circ\)
1. \(45^\circ\)
2. \(55^\circ\)
3. \(35^\circ~\)
4. \(40^\circ\)
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In the first excited state of hydrogen atom, the energy of its electron is \(-3.4~\text{eV}.\) The radial distance of the electron from the hydrogen nucleus in this case is approximately:
(take \(1\text{ eV} = 1.6 \times 10^{-19}\text{ J}\), \(e = 1.6 \times 10^{-19}\text{ C}\) and \(\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2~\))
1. \(2.1 \times 10^{-8}\text{ m}~\)
2. \(2.1 \times 10^{-10}\text{ m}~\)
3. \(2.1 \times 10^{-11}\text{ m}~\)
4. \(2.1 \times 10^{-9}\text{ m}~\)
(take \(1\text{ eV} = 1.6 \times 10^{-19}\text{ J}\), \(e = 1.6 \times 10^{-19}\text{ C}\) and \(\dfrac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2/\text{C}^2~\))
1. \(2.1 \times 10^{-8}\text{ m}~\)
2. \(2.1 \times 10^{-10}\text{ m}~\)
3. \(2.1 \times 10^{-11}\text{ m}~\)
4. \(2.1 \times 10^{-9}\text{ m}~\)
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A uniform metallic wire having resistance \(4~\Omega\) is bent to form a square loop \((ABCD)\) (see figure). A resistance of \(2~\Omega\) is connected between points \(B\) and \(D\) and a battery of \(2~\text{V}\) is connected across points \(A\) and \(C\) as shown in the figure. Now the value of the current \((I)~\) is:
1. \(4~\text{A}\)
2. \(4.5~\text{A}~\)
3. \(8~\text{A}\)
4. \(2~\text{A}\)
1. \(4~\text{A}\)
2. \(4.5~\text{A}~\)
3. \(8~\text{A}\)
4. \(2~\text{A}\)
Level 4: Below 35%
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Match List-I with List-II.
Choose the correct answer from the options given below:
1. A-I, B-III, C-IV, D-II
2. A-III, B-IV, C-I, D-II
3. A-III, B-I, C-IV, D-II
4. A-IV, B-I, C-III, D-II
| List I (Complex/ ion) |
List II (Shape/geometry) |
||
| A. | \([Pt(Cl_2)(NH_3)_2]\) | I. | Octahedral |
| B. | \([Co(NH_3)_6]Cl_3\) | II. | Trigonal bipyramidal |
| C. | \([NiCl_4]^{2-}\) | III. | Square planar |
| D. | \([Fe(CO)_5]\) | IV. | Tetrahedral |
Choose the correct answer from the options given below:
1. A-I, B-III, C-IV, D-II
2. A-III, B-IV, C-I, D-II
3. A-III, B-I, C-IV, D-II
4. A-IV, B-I, C-III, D-II
Level 4: Below 35%
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Calculate emf of the half cell given below :
Pt (s) | \(H_2\) (g, 2 atm) | HCl (aq, 0.02 M)
\(E^0_{H_2 / H^+}\) = 0 V
(Given : \(\dfrac{2.303 RT}{F} = 0.059~\),
log 2 = 0.3010)
1. − 0.109 V
2. 0.109 V
3. 0.035 V
4. − 0.035 V
Pt (s) | \(H_2\) (g, 2 atm) | HCl (aq, 0.02 M)
\(E^0_{H_2 / H^+}\) = 0 V
(Given : \(\dfrac{2.303 RT}{F} = 0.059~\),
log 2 = 0.3010)
1. − 0.109 V
2. 0.109 V
3. 0.035 V
4. − 0.035 V
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At 298 K, a certain buffer solution contains equal concentrations of X- and HX, Kb for X- is 10-10.
What is the pH of this buffer solution ?
1. 10
2. 4
3. 2
4. 6
What is the pH of this buffer solution ?
1. 10
2. 4
3. 2
4. 6
Level 4: Below 35%
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Given below are certain reactions. Identify the reaction for which \(K_p \neq K_c.~\)
1. \( \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g}) ~\)
2. \( \mathrm{H}_2 \mathrm{O}(\mathrm{~g})+\mathrm{CO}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) ~\)
3. \( \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ~\)
4. \( \mathrm{H}_2(\mathrm{~g})+1_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})~\)
1. \( \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{~g}) ~\)
2. \( \mathrm{H}_2 \mathrm{O}(\mathrm{~g})+\mathrm{CO}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) ~\)
3. \( \mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ~\)
4. \( \mathrm{H}_2(\mathrm{~g})+1_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{~g})~\)
Level 4: Below 35%
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For a certain reaction R → Product, the plot of concentration [R] vs time has a negative slope as shown. The order of reaction is :
1. 1
2. 2.5
3. 2
4. 0
1. 1
2. 2.5
3. 2
4. 0
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