Sample QuestionsJEE Main 23-Jan-2025 Paper - Shift 1 questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
One die has two faces marked 1 , two faces marked 2 , one face marked 3 and one face marked 4. Another die has one face marked 1 , two faces marked 2 , two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5 , when both the dice are thrown together, is
- ✓
$\frac{1}{2}$
- B
$\frac{3}{5}$
- C
$\frac{2}{3}$
- D
$\frac{4}{9}$
Answer: A.
View full solution →If the system of equations
$
\begin{array}{l}
(\lambda-1) x+(\lambda-4) y+\lambda z=5 \\
\lambda x+(\lambda-1) y+(\lambda-4) z=7 \\
(\lambda+1) x+(\lambda+2) y-(\lambda+2) z=9
\end{array}
$
has infinitely many solutions, then $\lambda^2+\lambda$ is equal to
Answer: B.
View full solution →If $A , B$ and $\left(\operatorname{adj}\left( A ^{-1}\right)+\operatorname{adj}\left( B ^{-1}\right)\right)$ are non-singular matrices of same order, then the inverse of $A \left(\operatorname{adj}\left( A ^{-1}\right)+\operatorname{adj}\left( B ^{-1}\right)\right)^{-1} B$, is equal to
- A
$A B^{-1}+A^{-1} B$
- B
$\operatorname{adj}\left( B ^{-1}\right)+\operatorname{adj}\left( A ^{-1}\right)$
- ✓
$\frac{1}{| AB |}(\operatorname{adj}( B )+\operatorname{adj}( A ))$
- D
$\frac{ AB ^{-1}}{|A|}+\frac{ BA ^{-1}}{|B|}$
Answer: C.
View full solution →Let the position vectors of the vertices $A , B$ and C of a tetrahedron ABCD be $\hat{ i }+2 \hat{ j }+\hat{ k }, \hat{ i }+3 \hat{ j }-2 \hat{ k }$ and $2 \hat{i}+\hat{j}-\hat{k}$ respectively. The altitude from the vertex $D$ to the opposite face ABC meets the median line segment through A of the triangle $A B C$ at the point $E$. If the length of $A D$ is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6 \sqrt{2}}$, then the position vector of E is
- A
$\frac{1}{2}(\hat{ i }+4 \hat{ j }+7 \hat{ k })$
- B
$\frac{1}{12}(7 \hat{ i }+4 \hat{ j }+3 \hat{ k })$
- C
$\frac{1}{6}(12 \hat{i}+12 \hat{j}+\hat{k})$
- ✓
$\frac{1}{6}(7 \hat{ i }+12 \hat{ j }+\hat{ k })$
Answer: D.
View full solution →Marks obtains by all the students of class 12 are presented in a frequency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of students whose marks are less than 12 is 18 , then the total number of students is
Answer: B.
View full solution →If the equation $a(b-c) x^2+b(c-a) x+c(a-b)=0$ has equal roots, where $a + c =15$ and $b =\frac{36}{5}$, then $a ^2+ c ^2$ is equal to __________.
View full solution →If the set of all values of a, for which the equation $5 x^3-15 x-a=0$ has three distinct real roots, is the interval $(\alpha, \beta)$, then $\beta-2 \alpha$ is equal to __________.
View full solution →Let the circle $C$ touch the line $x-y+1=0$, have the centre on the positive $x$-axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3 x+2 y=1$. Let H be the hyperbola $\frac{ x ^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C . Then $2 \alpha^2+3 \beta^2$ is equal to __________.
View full solution →The sum of all rational terms in the expansion of $\left(1+2^{1 / 3}+3^{1 / 2}\right)^6$ is equal to __________.
View full solution →If the area of the larger portion bounded between the curves $x^2+y^2=25$ and $y=|x-1|$ is $\frac{1}{4}(b \pi+c)$, $b , c \in N$, then $b + c$ is equal to $\qquad$
View full solution →Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of a radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be-
Answer: D.
View full solution →Answer: D.
View full solution →The electric field of an electromagnetic wave in free space is
$
\overrightarrow{E}=57 \cos \left[7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)\right]
$
$
(4 \hat{i}-3 \hat{j}) N / C
$
The associated magnetic field in Tesla is-
- A
$\overrightarrow{ B }=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t -5 \times 10^{-3}(3 x +4 y )\right](5 \hat{ k })$
- B
$\overrightarrow{ B }=\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t -5 \times 10^{-3}(3 x +4 y )\right](\hat{ k })$
- ✓
$\overrightarrow{ B }=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t -5 \times 10^{-3}(3 x +4 y )\right](5 \hat{ k })$
- D
$\overrightarrow{ B }=-\frac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t -5 \times 10^{-3}(3 x +4 y )\right]$ $(\hat{ k })$
Answer: C.
View full solution →Given a thin convex lens (refractive index $\mu_2$ ), kept in a liquid (refractive index $\mu_1, \mu_1<\mu_2$ ) having radii of curvature $\left|R_1\right|$ and $\left|R_2\right|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?
- A
$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(\left| R _1\right|+\left| R _2\right|\right)-\mu_1\left| R _1\right|}$
- ✓
$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(\left| R _1\right|+\left| R _2\right|\right)-\mu_1\left| R _2\right|}$
- C
$\frac{\mu_1\left| R _1\right| \cdot\left| R _2\right|}{\mu_2\left(2\left| R _1\right|+\left| R _2\right|\right)-\mu_1 \sqrt{\left| R _1\right| \cdot\left| R _2\right|}}$
- D
$\frac{\left(\mu_2+\mu_1\right)\left|R_1\right|}{\left(\mu_2-\mu_1\right)}$
Answer: B.
View full solution →The electric flux is $\phi=\alpha \sigma+\beta \lambda$ where $\lambda$ and $\sigma$ are linear and surface charge density, respectively, $\left(\frac{\alpha}{\beta}\right)$ represents
Answer: C.
View full solution →Two particles are located at equal distance from origin. The position vectors of those are represented by $\overrightarrow{ A }=2 \hat{ i }+3 n \hat{ j }+2 \hat{ k }$ and $\vec{B}=2 \hat{i}-2 \hat{j}+4 p \hat{k}$, respectively. If both the vectors are at right angle to each other, the value of $n ^{-1}$ is__________.
View full solution →A positive ion A and a negative ion B has charges $6.67 \times 10^{-19} C$ and $9.6 \times 10^{-10} C$, and masses $19.2 \times 10^{-27} kg$ and $9 \times 10^{-27} kg$ respectively. At an instant, the ions are separated by a certain distance r. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is $P \times 10^{-13}$, where the value of $P$ is __________ .
(Take $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 Nm ^2 C ^{-1}$ and universal gravitational constant as $6.67 \times 10^{-11} Nm ^2 kg^{-2}$ )
View full solution →Which among the following react with Hinsberg's reagent?
Choose the correct answer from the options given below :
Answer: D.
View full solution →Match the list - i with list - ii
LIST - I
(Classification of molecules,based on octet rule) | LIST - II
(Example) |
| A. | Molecules obeying octet rule | I. | NO,NO_(2) |
| B. | Molecules with incomplete octet | II. | BCl_(3),AlCl_(3) |
| C. | Molecules with incomplete octet with odd electron | III. | H_(2)SO_(4),PCl_(5) |
| D. | Molecules with expanded octet | IV. | CCl_(4),CO_(2) |
choose the correct answer from the questions given below :
Answer: A.
View full solution →Which of the following happens when $NH _4 OH$ is added gradually to the solution containing $1 M A ^{2+}$ and $1 M B ^{3+}$ ions ?
Given : $K _{\text {sp }}\left[ A ( OH )_2\right]=9 \times 10^{-10}$ and
$
K_{\text {sp }}\left[B(OH)_3\right]=27 \times 10^{-18} \text { at } 298 K
$
- ✓
$B ( OH )_3$ will precipitate before $A ( OH )_2$
- B
$A ( OH )_2$ and $B ( OH )_3$ will precipitate together
- C
$A ( OH )_2$ will precipitate before $B ( OH )_3$
- D
Both $A ( OH )_2$ and $B ( OH )_3$ do not show precipitation with $NH _4 OH$
Answer: A.
View full solution →Given below are two statements :
Statement I : In Lassaigne's test, the covalent organic molecules are transformed into ionic compounds.
Statement II : The sodium fusion extract of an organic compound having N and S gives prussian blue colour with $FeSO _4$ and $Na _4\left[ Fe ( CN )_6\right]$In the light of the above statements, choose the correct answer from the options given below
- A
Both Statement I and Statement II are true
- B
Both Statement I and Statement II are false
- C
Statement I is false but Statement II is true
- ✓
Statement I is true but Statement II is false
Answer: D.
View full solution →The correct set of ions (aqueous solution) with same colour from the following is :
- ✓
$V ^{2+}, Cr ^{3+}, Mn ^{3+}$
- B
$Zn ^{2+}, V ^{3+}, Fe ^{3+}$
- C
$Ti ^{4+}, V ^{4+}, Mn ^{2+}$
- D
$Sc ^{3+}, Ti ^{3+}, Cr ^{2+}$
Answer: A.
View full solution →The standard enthalpy and standard entropy of decomposition of $N _2 O _4$ to $NO _2$ are $55.0 kJ mol ^{-1}$ and $175.0 J / K / mol$ respectively. The standard free energy change for this reaction at $25^{\circ} C$ in $J mol ^{-1}$ is __________ (Nearest integer)
View full solution →For the thermal decomposition of $N _2 O _5(g)$ at constant volume, the following table can be formed, for the reaction mentioned below :
$
2 N_2 O_5(g) \rightarrow 2 N_2 O_4(g)+O_2(g)
$| S.No. | Time/s | Total pressure / (atm) |
| 1 | 0 | 0.6 |
| 2 | 100 | ('X') |
$x=$ __________ $\times 10^{-3} atm$ [nearest integer] Given : Rate constant for the reaction is $4.606 \times 10^{-2} s^{-1}$. View full solution →Consider the following sequence of reactions to produce major product $( A )$
Molar mass of product $( A )$ is __________ $g mol ^{-1}$.
(Given molar mass in $g mol ^{-1}$ of $C : 12, H : 1$,
$
O: 16, Br: 80, N: 14, P: 31)
$ View full solution →During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is __________ %.
(Given molar mass in $g mol ^{-1}$ of $Ba : 137, S: 32$, O:16)
View full solution →If 1 mM solution of ethylamine produces $pH =9$, then the ionization constant $\left( K _{ b }\right)$ of ethylamine is $10^{-x}$. The value of $x$ is _________ (nearest integer).
[The degree of ionization of ethylamine can be neglected with respect to unity.]
View full solution →