UPSEE 2017 Question Paper with detailed Solutions and Answer Key Released, check here
The first shift of Uttar Pradesh State Examination (UPSEE/UPTU) 2017 Examination was held on April 16, 2017. The question paper with its detailed solution and answer key has been released on the official website of UPSEE. According to the feedback of students, the difficulty level of this paper was moderate Students can download UPSEE/UPTU 2017 Question Paper and its detailed solution from this article. This paper will be very useful for those who are going to appear for the exam on 22 and 23 April 2017.
Few questions from the Question Paper are given below:
Q. A light beam consists of two types of photons. In one type each photon has energy 2eV and in other type each photon has energy 3eV. The light beam is incident on a photoelectric material of work function 1eV. The maximum kinetic energy of emitted photoelectron is :
Correct Option: (B)
KE max = hνmax - Φ = 3 - 1 = 2 eV
Q. Gravitational force acts on a particle due to fixed uniform solid sphere. Neglect other forces. Then particle :
(A) experiences a force directed along the radial direction only.
(B) always moves normal to the radial direction
(C) always moves in the radial direction only.
(D) always moves in circular orbit.
Correct Option: (A)
Circular motion is a special case in gravitational field. There may be straight line,elliptical paths, but force will be always directed toward the centre of the sphere.
Q. Which of the following statement is correct ?
(A) BCl3 and AlCl3 are both Lewis acids and BCl3 is stronger than AlCl3
(B) BCl3 and AlCl3 are both Lewis acids and AlCl3 is stronger than BCl3
(C) BCl3 and AlCl3 are both equally strong Lewis acid
(D) Both BCl3 and AlCl3 are not Lewis acids.
Correct Option: (A)
BCl3 and AlCl3 both are Lewis acids but BCl3 is more electron deficient than AlCl3 so BCl3 is stronger Lewis acid than AlCl3 due to high electron negativity.(E.N. B-2.0,Al 1.5)
Q. Given y = x2 . As x→2, y→4 what must the value of δ be for which from | x – 2 |< δ it follows that
| y – 4 | < ∈ = 0.001 ?
(A) 0 < δ < 0.00025
(B) 0.03 < δ < 0.05
(C) 0.2 < δ < 0.25
(D) 0.4< δ < 0.5
| x – 2 |< δ, we get | y – 4 | < δ(δ + 4), which is less than ∈
So, δ < √(∈ + 4) - 2
For , ∈ = 0.001, the δ < 0.00025