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AP ECET 2017 Electronics & Instrumentation Engineering Question Paper with Answer Key pdf is available for download. The exam was conducted by Jawaharlal Nehru Technological University, Ananthapur on May 3, 2017 in the Forenoon Session 10 AM to 1 PM. The question paper comprised a total of 200 questions.
AP ECET 2017 Electronics & Instrumentation Engineering Question Paper with Answer Key PDF
| AP ECET 2017 Electronics & Instrumentation Engineering Question Paper PDF | AP ECET 2017 Electronics & Instrumentation Engineering Answer Key PDF |
|---|---|
| Download PDF | Download PDF |
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AP ECET Questions
1. If $I(x)=\int x^{2}(\log x)^{2} d x$ and $I( 1)=0$, then $I(x)$
If $I(x)=\int x^{2}(\log x)^{2} d x$ and $I( 1)=0$, then $I(x)$
- $\frac{x^{3}}{18} \left[ 8\left(\log x\right)^{2} -3\log x\right] + \frac{7}{18} $
- $\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} +6 \log x\right] - \frac{2}{27} $
- $\frac{x^{3}}{27} \left[9\left(\log x\right)^{2} - 6 \log x+2\right]- \frac{2}{27} $
- $\frac{x^{3}}{27} \left[9 \left(\log x\right)^{2} -6 \log x +2 \right] - \frac{2}{27}$
2. A parallel plate capacitor consists of two circular plates each of radius $2 \,cm$, separated by a distance of $0.1 \,mm$. If the potential difference across the plates is varying at the rate of $5 \times 10^{6} Vs ^{-1}$, then the value of displacement current is
A parallel plate capacitor consists of two circular plates each of radius $2 \,cm$, separated by a distance of $0.1 \,mm$. If the potential difference across the plates is varying at the rate of $5 \times 10^{6} Vs ^{-1}$, then the value of displacement current is
- 5.56 A
- 5.56 mA
- 0.556 mA
- 2.28 mA
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. The rate constants for a reaction at $400\, K$ and $500\, K$ are $2.60 \times 10^{-5} s ^{-1}$ and $2.60 \times 10^{-3} s ^{-1}$
respectively. The activation energy of the reaction in $kJ\, mol ^{-1}$ is
The rate constants for a reaction at $400\, K$ and $500\, K$ are $2.60 \times 10^{-5} s ^{-1}$ and $2.60 \times 10^{-3} s ^{-1}$
respectively. The activation energy of the reaction in $kJ\, mol ^{-1}$ is
- 38.3
- 57.4
- 114.9
- 76.6
5. Magnesium is burnt in air to form $A$ and $B$. When $B$ is hydrolysed, $C$ and $D$ are formed. $D$ is the reactant in the manufacture of nitric acid by Ostwald's process. What is $C$
Magnesium is burnt in air to form $A$ and $B$. When $B$ is hydrolysed, $C$ and $D$ are formed. $D$ is the reactant in the manufacture of nitric acid by Ostwald's process. What is $C$
- $\ce{NH_3}$
- $\ce{Mg(OH)_2}$
- $MgO$
- $NO$
6. 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
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