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TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

TS EAMCET 2025 May 4 Shift 1
Chemistry

121) The radius of second orbit of hydrogen atom is same as that of orbit (n) of an ion (x). n and x are respectively

A) 4, Be²⁺

B) 3, Li²⁺

C) 4, Be³⁺

D) 2, He⁺

View Answer

C) 4, Be³⁺
Explanation:The radius of the n-th orbit in Bohr’s model is given by the formula $r_n = a_0 \frac{n^2}{Z}$

.
For a hydrogen atom, the atomic number Z = 1 and the second orbit means n = 2.
The radius of the second orbit of hydrogen is $r_{\text{H}} = a_0 \frac{2^2}{1} = 4a_0$

.
For the ion (x) in orbit n, the radius is $r_x = a_0 \frac{n^2}{Z}$

.
Setting the radii equal, we get $4a_0 = a_0 \frac{n^2}{Z}$

, which simplifies to n² = 4Z.
We now check the options for hydrogen-like species (ions containing only one electron):
For Be³⁺, the atomic number Z = 4.
Substituting Z = 4 into the equation: n² = 4 × 4 = 16.
Taking the square root, we find n = 4.
This confirms that for n = 4 and x = Be³⁺, the radius matches that of the second orbit of hydrogen.

122) An electromagnetic radiation of wavelength 331.5 nm is made to strike the surface of a metal. Electrons are emitted with a kinetic energy of $1.2 \times 10^5 \text{ J mol}^{-1}$ . The work function (in eV) of the metal is ( $h = 6.63 \times 10^{-34} \text{ Js}$ , $N_A = 6 \times 10^{23} \text{ mol}^{-1}$ )

A) 1.5

B) 3.0

C) 3.5

D) 2.5

View Answer

D) 2.5
Explanation:Given:
λ = 331.5 nm = 3.315 × 10⁻⁷ m
KE = 1.2×10⁵ J/mol
Energy per electron:
= (1.2×10⁵) / (6×10²³)
= 2 × 10⁻¹⁹ J
Photon energy:
E = hc/λ
= (6.63×10⁻³⁴ × 3×10⁸) / (3.315×10⁻⁷)
≈ 6 × 10⁻¹⁹ J
Work function:
Φ = E − KE
= (6 − 2) ×10⁻¹⁹ = 4×10⁻¹⁹ J
Convert to eV:
Φ = (4×10⁻¹⁹)/(1.6×10⁻¹⁹)
= 2.5 eV

123) Match elements with block

List-1 (Element)		List-2 (Block)
A	Cd	I	f-block
B	Eu	II	s-block
C	Se	III	d-block
D	Ba	IV	p-block

A) A-IV, B-III, C-II, D-I

B) A-II, B-IV, C-I, D-III

C) A-III, B-IV, C-II, D-I

D) A-III, B-I, C-IV, D-II

View Answer

D) A-III, B-I, C-IV, D-II
Explanation:Block classification

124) In long form of periodic table an element ‘E’ has atomic number 78.The period and group number of the element are x and y respectively. (x+y) is equal to

A) 18

B) 15

C) 17

D) 16

View Answer

D) 16
Explanation:Pt ⇒ group 10, period 6 ⇒ sum 16

125) In which of the following options, the molecules are correctly arranged in the increasing order of their bond angles?

A) NH₃ < O₃ < H₂O < SO₂

B) H₂O < O₃ < NH₃ < SO₂

C) H₂O < NH₃ < SO₂ < O₃

D) H₂O < NH₃ < O₃ < SO₂

View Answer

D) H₂O < NH₃ < O₃ < SO₂
Explanation:Lone pair repulsion ⇒ order

126) In which of the following, the compounds are correctly arranged in the de creasing order of boiling points?

A) HF > H₂O > NH₃ > PH₃

B) H₂O > HF > NH₃ > PH₃

C) H₂O > HF > PH₃ > NH₃

D) HF > NH₃ > H₂O > PH₃

View Answer

B) H₂O > HF > NH₃ > PH₃
Explanation:Hydrogen bonding order

127) The force (F) required to maintain the flow of layers of a liquid is equal to (A = area of contact of layers, dz = distance between the layers, du = change in velocity, η = coefficient of viscosity)

A) $\eta \frac{du}{dz} \frac{1}{A}$

B) $\eta \frac{dz}{du}A$

C) $\eta A\frac{du}{dz}$

D) $\eta\frac{dz}A\frac1{du}$

View Answer

C) $\eta A\frac{du}{dz}$

Explanation: $F=\eta A \frac{du}{dz}$

128) Consider the following redox reaction in basic medium.
$x\text{Cr(OH)}_3 + y(\text{IO}_3)^- + z(\text{OH})^- \rightarrow a(\text{CrO}_4)^{2-} + b(\text{I})^- + c(\text{H}_2\text{O})$ .
The incorrect option about it is

A) x+y=3

B) a+b=7

C) z=4

D) b=1

View Answer

B) a+b=7
Explanation:Oxidation:
Cr(OH)₃ → CrO₄²⁻ (Cr: +3 → +6, loss = 3e⁻)
Reduction:
IO₃⁻ → I⁻ (I: +5 → −1, gain = 6e⁻)
LCM of electrons = 6
⇒ 2Cr(OH)₃ + IO₃⁻ + OH⁻ → 2CrO₄²⁻ + I⁻ + H₂O
Balanced equation:
2Cr(OH)₃ + IO₃⁻ + 4OH⁻ → 2CrO₄²⁻ + I⁻ + 5H₂O
So:
x=2, y=1, z=4, a=2, b=1
Check options:
x+y = 3 ✔
a+b = 3 ✘
z = 4 ✔
b = 1 ✔

129) The entropy and enthalpy changes for the reaction $CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)$ at 300 K and 1 atm are respectively $-42.4 \text{ JK}^{-1}$ and $-41.2 \text{ kJ}$ . The temperature at which the reaction will go in the reverse direction is

A) 761.8 K

B) 671.8 K

C) 961.8 K

D) 971.8 K

View Answer

D) 971.8 K
Explanation:ΔS = −42.4 J/K = −0.0424 kJ/K
ΔH = −41.2 kJ
At equilibrium:
ΔG = 0 ⇒ T = ΔH / ΔS
T = (−41.2)/(−0.0424) ≈ 971.8 K

130) The volume of water required to dissolve 0.1 g $PbCl_2$ to get a saturated solution (in mL) is (Given $K_{sp}(PbCl_2) = 3.2 \times 10^{-8}$ ; Atomic mass of Pb = 207u)

A) 150

B) 100

C) 120

D) 180

View Answer

D) 180
Explanation:PbCl₂ ⇌ Pb²⁺ + 2Cl⁻
Ksp = 4s³ = 3.2×10⁻⁸
⇒ s³ = 0.8×10⁻⁸
⇒ s = 2×10⁻³ mol/L
Molar mass PbCl₂ ≈ 278 g/mol
Solubility in g/L:
= 278 × 2×10⁻³ = 0.556 g/L
Volume required for 0.1 g:
= 0.1 / 0.556 ≈ 0.18 L = 180 mL

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TS EAMCET 2025 May 4 Shift 1 Chemistry

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TS EAMCET 2025 May 4 Shift 1
Chemistry