Chloride radical is detected by the chromyl chloride test. In this test, chromyl chloride gas (orange red color) is produced. Equation Involved – 4NaCl + K2Cr2O7 + 6H2SO4 → 4NaHSO4 + 2KHSO4 + 3H2O +...
The lanthanide contraction is the decrease in the atomic or ionic radii with increase in the atomic number of lanthanides
Lanthanide elements resembles a lot in properties with lanthanum. Lanthanide is group of 14 elements from atomic number 58 to 71. In these elements on increasing atomic number electron enters into...
2KMnO4 + 8H2SO4 + 10KI → 6K2SO4 + 8H2O + 5I2 2KMnO4 + 5SO2 + 2H2O → 2MnSO4 + 2H2SO4 + K2SO4 2KmO4 + 16HCl → 2KCl + 2MnCl2 + 8H2O + 5Cl2 5COOH – COOH + [5O] → 10CO2 + 5H2O
Answer: Double salt Complex salt A double salt is a combination of two salt compounds. A complex salt is a molecular structure that is composed of one or more complex ions. Double salts can give...
Answer: DNA RNA DNA – Deoxyribo Nucleic Acid RNA – Ribo Nucleic acid DNA consists of adenine (A), cytosine (C), guanine (G), and thymine (T) RNA consists of adenine (A), cytosine (C), guanine (G),...
![\[\begin{array}{l} \square \vec{r}=(\hat{i}+\hat{j})+\lambda(2 \hat{i}-\hat{j}+\hat{k}) \\ \vec{r}=(2 \hat{i}+\hat{j}-\hat{k})+\mu(3 \hat{i}-5 \hat{j}+2 \hat{k}) \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2f8ab5b81e582387cb9ae472791476e4_l3.png "Rendered by QuickLaTeX.com")
Answer Given equations: r¯=(ı^+ȷ^)+λ(2ı^-ȷ^+k^) \overline{\mathrm{r}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{\mathrm{k}})...
![\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{-2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c5a91864ba7e1e0ed1b5d273d822f5e5_l3.png "Rendered by QuickLaTeX.com")
(b) ![\left[\mathrm{Pd}(\mathrm{CN})_{4}\right]^{2-}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ffda616d3dbe7dc8925722e87005924e_l3.png "Rendered by QuickLaTeX.com")
(c) ![\left[\mathrm{PdCl}_{4}\right]^{2-}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-90cc6a4f4d42dee46e06d6f8ed516b7a_l3.png "Rendered by QuickLaTeX.com")
(d) ![\left[\mathrm{NiCl}_{4}\right]^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-149ee5c95bd9cfcc4405704fba2e1bab_l3.png "Rendered by QuickLaTeX.com")
SOL: Correct option is D. $\left[\mathrm{NiCl}_{4}\right]^{2}$
(d) all of theseSol: Correct option is B. 0
(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...
(a) Both Assertion (A) and reason (R) are true and Reason (R) is a correct explanation of Assertion (A). (b) Both Assertion (A) and reason (R) are true and Reason (R) is not a correct explanation of...
(a) 13 (b) 14 (c) 15 (d) 16 Answer: (c) 15 Sol: Median of 6 numbers is the average of 3rd and 4th term. ∴ 13 = (????−1)+(????−3) 2 ⇒ 26 = 2x – 4 ⇒ 2x = 30 ⇒ x = 15 Thus, x is equal to...
(a) 0 (b) 1 (c) 10 (d) 19 Answer: (d) 19 Sol: It is given that mean of 20 numbers is zero. i.e., average of 20 numbers is zero. i.e., sum of 20 numbers is zero. Thus, at most, there can be 19...
(a) 7 (b) 9 (c) 11 (d) 13 Answer: (b) 9 Sol: First 8 prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19. Median of 8 numbers is average of 4th and 5th terms.
Answer: (c) 13 Sol: Converting the given data into a frequency table, we get: Hence, the number of families having an income range of Rs. 20,000 – Rs. 25,000 is 13. The correct option is...
(a) mean ˂ mode ˂ median (b) mean > mode > median (c) mean = mode = median (d) mode = 1 2 (mean + median) Answer: (c) mean = mode = median Sol: A symmetric distribution is one where the left...
(a) 27.5 (b) 24.5 (c) 28.4 (d) 25.8 Answer: (b) 24.5 Sol: Mode = (3 × median) – (2 × mean) ⇒ (2 × mean) = (3 × median) – mode ⇒ (2 × mean) = 3 × 26 – 29 ⇒ (2 × mean) = 49 ⇒ Mean = 49 2 ∴ Mean = 24.5...
(a) 22 (b) 23.5 (c) 24 (d) 24.5 Answer: (c) 24 Sol: Mode = (3 × median) – (2 × mean) ⇒ (3 × median) = (mode + 2 mean) ⇒ (3 × median) = 16 + 56 ⇒ (3 × median) = 72 ⇒ Median = 72 3 ∴...
(a) 7.2 (b) 8.2 (c) 9.2 (d) 10.2 Answer: (c) 9.2 Sol: It is given that the mean and median are 8.9 and 9, respectively, ∴ Mode = (3 × Median) – (2 × Mean) ⇒ Mode = (3 × 9) – (2 × 8.9) = 27 – 17.8 =...
(a) 5.5 (b) 15.5 (c) 20.5 (d) 36.0 Answer: (c) 20.5 Sol: The x- coordinate represents the median of the given data. Thus, median of the given data is 20.5.
(a) mode=(3 * mean) – (2 * median) (b) mode=(3 *median) – (2 *mean) (c) median=(3 * mean) – (2 * mode) (d) mean=(3 * median) – (2 *mode) Answer: (b) mode=(3 * median) – (2 *mean) Sol: mode=(3 *...
(a) evenly distributed over the classes (b) centred at the class marks of the classes (c) centred at the lower limits of the classes (d) centred at the upper limits of the classes Answer: (b)...
(a) lower limits of the classes (b) upper limits of the classes (c) midpoints of the classes (d) none of these Answer: (c) midpoints of the classes Sol: The ???????? ′???? are the deviations from A...
(a) Mean (b) Median (c) Mode (d)None of these Answer: (b) Median Sol: The abscissa of the point of intersection of the ‘less than type’ and that of the ‘more than type’ cumulative...
(a) Mean (b) Median (c) Mode (d) all of these Answer: (b) Median Sol: The cumulative frequency table is useful in determining the median.
(a) a histogram (b) a frequency curve (c) a frequency polygon (d) ogives Answer: (d) ogives Sol: This because median of a frequency distribution is found graphically...
(a) a frequency curve (b) a frequency polygon (c) a histogram (d) an ogive Answer: (c) a histogram Sol: The mode of a frequency distribution can be obtained graphically from a histogram.
(a) Mean (b) Median (c) Mode (d) None of these Answer: (a) Mean Sol: Mean is influenced by extreme values.
(a) Mean (b) Median (c) Mode (d) None of these Answer: (a) Mean Sol: The mean cannot be determined graphically because the values cannot be summed.
(a) Mean (b) Mode (c) Median (d) Standard Deviation Answer: (d) Standard Deviation Sol: The standard deviation is a measure of dispersion. It is the action or process of distributing...
Sol: Since, 8 is less than 30 and 32 is more than 30, so the middle value remains unchanged Thus, the median of 21 observations taken together is 30.
Sol: Upper class boundary = Lowest class boundary + width × number of classes = 8.1 + 2.5×12 = 8.1 + 30 = 38.1 Thus, upper class boundary of the highest class is 38.1.
and 
are collinear then the value of 
is 
Solution: Option(C) To find: Area of $A B C$ Given: $A(3,-2), B(k, 2)$ and $C(8,8)$ The formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} &...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8\end{array}\right|=0$ Applying $\mathrm{R}_{1}...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0$ Applying...
is 
Solution: Option(A) To find: Value of $x$ We have, $\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ x-4 & 2 x-9 & 3 x-16 \\ x-8 & 2 x-27 & 3 x-64\end{array}\right|=0$ Applying...
then 
Solution: Option(C) To find: Value of $\mathrm{x}$ We have, $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & 2 \\ 9 & 6 & -2\end{array}\right|=0$ Applying $R_{1} \rightarrow 2...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}a & 1 & b+c \\ b & 1 & c+a \\ c & 1 & a+b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & 1 &...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{a}+\mathrm{c} & 1 \\ \mathrm{ab} &...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}b+c & a & b \\ c+a & c & a \\ a+b & b & c\end{array}\right|$ We have,...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$ We...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$ We have,...
be distinct positive real numbers then the value of 
is Solution: Option(B) To find: Nature of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & b & c \\...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q\end{array}\right|$ We have,...
Solution: Option(D) To find: Value of $\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ We have, $\left|\begin{array}{lll}a-b...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$ We have,...
is a complex root of unity then 
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|$ We have,...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\sin 23^{\circ} & -\sin 7^{\circ} \\ \cos 23^{\circ} & \cos 7^{\circ}\end{array}\right|$ Formula used: (i) $\sin (A+B)=\sin A...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\cos 70^{\circ} & \sin 20^{\circ} \\ \sin 70^{\circ} & \cos 20^{\circ}\end{array}\right|$ Formula used: (i) $\cos \theta=\sin...
and 
are collinear, prove that 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
be three points such that area of a 
is 4 sq units, find the value of .Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the area of a ABC having vertices 
and 
is 35 sq units.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points and are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: Operating $\mathrm{C}_{1} \rightarrow \mathrm{aCl}_{1}$ $=\frac{1}{a}\left|\begin{array}{ccc} a^{2} & b-c & c+b \\ a^{2}+a c & b & c-a \\ a^{2}-a b & b+a & c...
Solution: Operating $R 1 \rightarrow R 1+R_{2}+R_{3}$ $0=\left|\begin{array}{ccc} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|$ Taking $(x+9)$ common from...
Solution: Operating $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $0=\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|$ Taking $(x+9)$ common...
Solution: $\text { Operating } \mathrm{Cl} \rightarrow \mathrm{Cl}+\mathrm{C}_{2}+\mathrm{C}_{3}$ $0=\left|\begin{array}{ccc} 3 x-8+3+3 & 3 & 3 \\ 3+3 x-8+3 & 3 x-8 & 3 \\ 3+3+3 x-8...
Solution: Operating $R 1 \rightarrow R 1^{-} R_{2}, R_{2 \rightarrow R 2}-R_{3}$ $\left|\begin{array}{ccc} 0 & x-b & x^{3}-b^{3} \\ 0 & b-c & b^{3}-c^{3} \\ 1 & c & c^{3}...
is a root of the equation 
.Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}$ $\begin{aligned} 0 &=\left|\begin{array}{ccc} x-2 & -6+3 x & -1-x+3 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x...
Solution: $\text { Operating } R 1 \rightarrow R 1^{-} R_{3}, R_{2 \rightarrow R_{2}-R 3}$ $\begin{array}{l} \left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{c} &...
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\rightarrow\left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{b} & \mathrm{b} c-\mathrm{ac} \\ 0 &...
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} 0 & \mathrm{a}^{2}+\mathrm{bc}-\mathrm{b}^{2}-\mathrm{ac} &...
and 
, prove that 
Solution: By properties of determinants, we can split the given determinant into 2 parts $\rightarrow 0=\left|\begin{array}{lll} \mathrm{x} & \mathrm{x}^{3} & \mathrm{x}^{4} \\ \mathrm{y}...
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & a+2-a-3 & 0 \\ (a+2)(a+3)-(a+3)(a+4)...
, where 
are in AP.Solution: Given that $\alpha, \beta, \gamma$ are in an $A P$, which means $2 \beta=\alpha+\gamma$ Operating $R_{3} \rightarrow R_{3}-2 R_{2}+R_{1}$ $\begin{array}{l} =\left|\begin{array}{ccc} x-3...
Solution: Taking $a, b, c$ from $C_{1}, C_{2}, C_{3}$ $=a b c\left|\begin{array}{ccc} -b^{2}-c^{2}+a^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & b^{2}-c^{2}-a^{2} & 2 c^{2} \\ 2 a^{2} &...
Sol: A line intersecting a circle at two district points is called a secant A circle can have two parallel tangents at the most The common point of a tangent to a circle and the circle is called the...
Answer: (d) A tangent to the circle can be drawn form a point inside the circle. Sol: A tangent to the circle can be drawn from a point Inside the circle. This statement is false because tangents...
Answer: (d) A straight line can meet a circle at one point only. Sol: A straight be can meet a circle at one point only This statement is not true because a straight line that is not a tangent but a...