Content Curator | Updated On - May 24, 2024
AP ECET 2024 Mining Engineering Question Paper is available for download here. JNTU Anantapur on behalf of APSCHE conducted AP ECET 2024 on May 8 Shift 2. AP ECET 2024 Mining Engineering Question Paper consists of 25 questions from Physics and Chemistry each, 50 questions from Mathematics and 100 questions from Mining Engineering to be attempted in the duration of 3 hours.
AP ECET 2024 Mining Engineering Question Paper with Answer Key PDF
| AP ECET 2024 Mining Engineering Question Paper | AP ECET 2024 Mining Engineering Answer Key |
|---|---|
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AP ECET Questions
1. A solution is prepared by dissolving $10 \,g$ of a non-volatile solute (molar mass, $'M^{\prime} g mol ^{-1}$ ) in $360\, g$ of water. What is the molar mass in $g\, mol ^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$ ?
A solution is prepared by dissolving $10 \,g$ of a non-volatile solute (molar mass, $'M^{\prime} g mol ^{-1}$ ) in $360\, g$ of water. What is the molar mass in $g\, mol ^{-1}$ of solute if the relative lowering of vapour pressure of solution is $5 \times 10^{-3}$ ?
- 199
- 99.5
- 299
- 149.5
2. A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200\, K$. It is suspended inside a chamber whose walls are at $0\, K$. The time required (in (is) for the temperature of the sphere to drop to $100 \,K$ is
($\sigma$ is Stefan's constant and all the quantities are in SI units)
A solid copper sphere of density $\rho$, specific heat capacity $C$ and radius $r$ is initially at $200\, K$. It is suspended inside a chamber whose walls are at $0\, K$. The time required (in (is) for the temperature of the sphere to drop to $100 \,K$ is
($\sigma$ is Stefan's constant and all the quantities are in SI units)
- $48 \frac{r\rho C}{\sigma}$
- $\frac{1}{48} \frac{r\rho C}{\sigma}$
- $\frac{27}{7} \frac{r\rho C}{\sigma}$
- $\frac{7}{27} \frac{r\rho C}{\sigma}$
3. A proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2\, mT$ perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by $90^{\circ}$. Then at this tune, the angle between the velocity vectors of proton and $\alpha$ - particle is
A proton and an $\alpha$-particle are simultaneously projected in opposite directions into a region of uniform magnetic field of $2\, mT$ perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by $90^{\circ}$. Then at this tune, the angle between the velocity vectors of proton and $\alpha$ - particle is
- $60^{\circ}$
- $90^{\circ}$
- $45^{\circ}$
- $180^{\circ}$
4. Let $A, G, H$ and $S$ respectively denote the arithmetic mean, geometric mean, harmonic mean and the sum of the numbers $a_1 , a_2 , a_3 ....., a_n$ . Then the value of at which the function $f(x) =\displaystyle \sum^n_{k =1} (x -a_k)^2$ has minimum is
Let $A, G, H$ and $S$ respectively denote the arithmetic mean, geometric mean, harmonic mean and the sum of the numbers $a_1 , a_2 , a_3 ....., a_n$ . Then the value of at which the function $f(x) =\displaystyle \sum^n_{k =1} (x -a_k)^2$ has minimum is
- S
- H
- G
- A
5. Let $f(x) = x^2 + 2x + 2, g(x) = - x^2 + 2x - 1 $ and $a, b$ be the extreme values of $f(x), g(x)$ respectively. If $c$ is the extreme value of $\frac{f}{g} (x)$ (for x $\neq$ 1), then $a + 2b + 5c + 4$ =
Let $f(x) = x^2 + 2x + 2, g(x) = - x^2 + 2x - 1 $ and $a, b$ be the extreme values of $f(x), g(x)$ respectively. If $c$ is the extreme value of $\frac{f}{g} (x)$ (for x $\neq$ 1), then $a + 2b + 5c + 4$ =
- 2
- 1
- 4
- 3
6. The half-life periods of a first order reaction at $300\, K$ and $400\, K$ are $50\, s$ and $10\, s$ respectively.
The activation energy of the reaction in $kJ \; mol^{-1}$ is (log 5 = 0.70)
The half-life periods of a first order reaction at $300\, K$ and $400\, K$ are $50\, s$ and $10\, s$ respectively.
The activation energy of the reaction in $kJ \; mol^{-1}$ is (log 5 = 0.70)
- 4
- 8
- 16.1
- 20.1
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