CBSE Study Material
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}})...
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
Answer Given - A = (1,2,3) B = (4,5,7) C = (-4,3,-6) D = (2,9,2) Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by ((a’-a),(b’-b),(c’-c))...
To prove – The lines are perpendicular to each other Direction ratios of L1 = (2,-2,1) Direction ratios of L2 = (2,1,-2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be...
To find – The value of λ Direction ratios of L1 = (-3,2λ,2) Direction ratios of L2 = (3λ,1,-5) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then...
To prove – The lines are perpendicular to each other Direction ratios of L1 = (2,-3,4) Direction ratios of L2 = (2,4,2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be...
To find – Angle between the two pair of lines Direction ratios of L1 = (-3,-2,0) Direction ratios of L2 = (1,-3,2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that...
To find – Angle between the two pair of lines Direction ratios of L1 = (1,0,-1) Direction ratios of L2 = (3,4,5) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
To find – Angle between the two pair of lines Direction ratios of L1 = (2,1,-3) Direction ratios of L2 = (3,2,-1) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
To find – Angle between the two pair of lines Direction ratios of L1 = (4,3,5) Direction ratios of L2 = (1,-1,1) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
To find – Angle between the two pair of lines Direction ratios of L1 = (1,1,2) Direction ratios of L2 = (3,5,4) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
To find – Angle between the two pair of lines Direction ratios of L1 = (2,-2,1) Direction ratios of L2 = (1,2,-2) Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of...
Tip – If (a,b,c) be the direction ratios of the first line and (a’,b’,c’) be that of the second, then the angle between these pair of lines is given by
It can thus be written as: A = (-4,2,-3) B = (1,3,-2) C = (-9,1,-4) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line...
Answer Given - A = (-1,4,-2) B = (λ,μ,1) C = (0,2,-1) To find – The value of λ and μ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios...
Answer Given - A = (3,2,-4) B = (9,8,-10) C = (λ,μ,-6) To find – The value of λ and μ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction...
Answer Given - A = (2,5,1) B = (1,2,-1) C = (3,λ,3) To find – The value of λ so that A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the...
Answer Given - A = (2,3,-4) B = (1,-2,3) C = (3,8,-11) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by...
Answer Given - A = (2,1,3) B = (5,0,5) C = (-4,3,-1) To prove – A, B and C are collinear Formula to be used – If P = (a,b,c) and Q = (a’,b’,c’),then the direction ratios of the line PQ is given by...
Find the image of the point $(2,-1,5)$ in the line r→=(11i^-2j^-8k^)+λ(10i^-4j^-11k^) \overrightarrow{\mathrm{r}}=(11 \hat{i}-2 \hat{j}-8 \hat{\mathrm{k}})+\lambda(10 \hat{i}-4...
Find the image of the point $(5,9,3)$ in the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}=2}{3}=\frac{\mathrm{z}-3}{4}$. Answer Given: Equation of line is...
Find the image of the point $(0,2,3)$ in the line $\frac{\mathrm{x}+3}{5}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}+4}{3}$. Answer Given: Equation of line is...
Find the coordinates of the foot of the perpendicular drawn from the point $A(1,8,4)$ to the line joining the points $B(0,-1,3)$ and $C(2,-3,-1)$ Answer Given: perpendicular drawn from point...
Find the coordinates of the foot of the perpendicular drawn from the point $A(1,2,1)$ to the line joining the points $B(1,4,6)$ and $C(5,4,4)$. Answer Given: perpendicular drawn from point...
Find the vector equation of a line passing through the point having the position vector $(\hat{i}+2 \hat{j}-3 \hat{k})$ and parallel to the line joining the points with position vectors...
Find the vector equation of a line passing through the point $A(3,-2,1)$ and parallel to the line joining the points $\mathrm{B}(-2,4,2)$ and $\mathrm{C}(2,3,3) .$ Also, find the Cartesian equations...
Find the vector and Cartesian equations of the line joining the points whose position vectors are $(\hat{i}-2 \hat{j}+\hat{k})$ and $(\hat{i}+3 \hat{j}-2 \hat{k})$ Answer Given: line passes through...
Find the vector and Cartesian equations of the line passing through the points $A(2,-3,0)$ and $B(-2,4,$, 3). Answer Given: line passes through the points $(2,-3,0)$ and $(-2,4,3)$ To find: equation...
Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-2}{-4}=\frac{\mathrm{Z}+7}{5}$. Also find the vector form of...
Find the coordinates of the foot of the perpendicular drawn from the point $(1,2,3)$ to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-7}{2}=\frac{\mathrm{z}-7}{-2}$. Also, find the length of the...
Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=z$ and $\frac{\mathrm{x}+1}{5}=\frac{\mathrm{y}-2}{1}, Z=2$ do not intersect each other. Answer Given: The equations of the two...
Show that the lines $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{Z}-3}{4}$ and $\frac{\mathrm{x}-4}{5}=\frac{\mathrm{y}-1}{2}=\mathrm{Z}$ intersect each other. Also, find the point...
Prove that the lines $\frac{\mathrm{x}-4}{1}=\frac{\mathrm{y}+3}{4}=\frac{\mathrm{z}+1}{7}$ and $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{-3}=\frac{\mathrm{z}+10}{8}$ intersect each other and find...
Find the Cartesian and vector equations of the line passing through the point $(1,2,-4)$ and perpendicular to each of the lines...
Find the equations of the line passing through the point $(-1,3,-2)$ and perpendicular to each of the lines $\frac{\mathrm{X}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{Z}}{3}$ and...
Find the Cartesian and vector equations of a line which passes through the point $(1,2,3)$ and is parallel to the line $\frac{-\mathrm{X}-2}{1}=\frac{\mathrm{y}+3}{7}=\frac{2 Z-6}{3}$ Answer Given:...
Find the equations of the line passing through the point $(1,-2,3)$ and parallel to the line $\frac{\mathrm{x}-6}{3}=\frac{\mathrm{y}-2}{-4}=\frac{\mathrm{Z}+7}{5}$. Also find the vector form of...
Find the Cartesian equations of the line which passes through the point $(1,3,-2)$ and is parallel to the line given by $\frac{\mathrm{x}+1}{3}=\frac{\mathrm{y}-4}{5}=\frac{\mathrm{z}+3}{-6}$. Also,...
The Cartesian equations of a line are $3 x+1=6 y-2=1-z$. Find the fixed point through which it passes, its direction ratios and also its vector equation. Answer Given: Cartesian equation of line are...
The Cartesian equations of a line are $\frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{Z}-6}{4}$. Find the vector equation of the line. Answer Given: Cartesian equation of line...
A line is drawn in the direction of $(\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})$ and it passes through a point with position vector $(2 \hat{i}-\hat{j}-4 \hat{k}) .$ Find the equations...
Find the vector equation of the line passing through the point with position vector $(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-5 \hat{\mathrm{k}})$ and parallel to the vector $(\hat{i}+3...
A line passes through the point $(2,1,-3)$ and is parallel to the vector $(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$. Find the equations of the line in vector and Cartesian forms....
Answer Given: line passes through point $(3,4,5)$ and is parallel to $2 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}$ To find: equation of line in vector and Cartesian forms Formula Used: Equation of a...
90C88 Answer : We know that: ⇒90C88=4005 Ans: ⇒90C88=4005
16c13 Answer : We know that: nCr = ⇒16C13= ⇒16C13= ⇒16C13 = ⇒16C13 = ⇒16C13 = ⇒16C13 =560 Ans: 16C13=560
21C4 Answer : We know that: nCr ⇒20C4 ⇒20C4 ⇒20C4 ⇒20C4 ⇒20C4 ⇒20C4 =4845 Ans: 20C4 =4845
(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...
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Answer: 1 ×10 1 = remainder obtained by dividing 1 × 1 by 10 = 1 3 ×10 7 = remainder obtained by dividing 3 × 7 by 10 = 1 7 ×10 9 = remainder obtained by dividing 7 × 9 by 10 = 3 Composition table:...
Answer: 1 ×5 1 = remainder obtained by dividing 1 × 1 by 5 = 1 3 ×5 4 = remainder obtained by dividing 3 × 4 by 5 = 2 4 ×5 4 = remainder obtained by dividing 4 × 4 by 5 = 1 Composition table: ×5 0 1...
Answer: 1 ×6 1 = remainder obtained by dividing 1 × 1 by 6 = 1 3 ×6 4 = remainder obtained by dividing 3 × 4 by 6 = 0 4 ×6 5 = remainder obtained by dividing 4 × 5 by 6 = 2 Composition table: ×6 0 1...
Answer: 1 +5 1 = remainder obtained by dividing 1 + 1 by 5 = 2 3 +5 1 = remainder obtained by dividing 3 + 1 by 5 = 2 4 +5 1 = remainder obtained by dividing 4 + 1 by 5 = 3 Composition Table: +5 0 1...
Answer: Given, ×4 on set S = {0, 1, 2, 3} 1 ×4 1 = remainder obtained by dividing 1 × 1 by 4 = 1 0 ×4 1 = remainder obtained by dividing 0 × 1 by 4 = 0 2 ×4 3 = remainder obtained by dividing 2 × 3...
Answers: (i) Consider, a, b ∈ Z a * b = a + b – 4 = b + a – 4 = b * a a * b = b * a, ∀ a, b ∈ Z Then, * is commutative on Z. a * (b * c) = a * (b + c – 4) = a + b + c -4 – 4 = a + b + c – 8 (a * b)...
Answer: Consider, a, b ∈ Q0 a * b = (3ab/5) = (3ba/5) = b * a a * b = b * a, for all a, b ∈ Q0 a * (b * c) = a * (3bc/5) = [a (3 bc/5)] /5 = 3 abc/25 (a * b) * c = (3 ab/5) * c = [(3 ab/5)...
Answers: (i) Consider, a, b ∈ Q – {-1} a * b = a + b + ab = b + a + ba = b * a a * b = b * a, ∀ a, b ∈ Q – {-1} a * (b * c) = a * (b + c + b c) = a + (b + c + b c) + a (b + c + b c) = a + b +...
Answer: Consider, a ∈ Q – {-1} and b ∈ Q – {-1} be the inverse of a. a * b = e = b * a a * b = e and b * a = e a + b + ab = 0 and b + a + ba = 0 b (1 + a) = – a Q – {-1} b = -a/1 + a Q – {-1}...
Answers: (i) Consider, X = (a, b) Y = (c, d) ∈ A, ∀ a, c ∈ R0 b, d ∈ R X O Y = (ac, bc + d) Y O X = (ca, da + b) X O Y = Y O X, ∀ X, Y ∈ A O is not commutative on A. X = (a, b) Y = (c, d) a Z = (e,...
Answer: Consider, F = (m, n) be the inverse in A ∀ m ∈ R0 and n ∈ R X O F = E F O X = E (am, bm + n) = (1, 0) and (ma, na + b) = (1, 0) Considering (am, bm + n) = (1, 0) am = 1 m = 1/a And bm + n =...
Answer: Consider, a ∈ Z and b ∈ Z be the inverse of a. a * b = e = b * a a * b = e and b * a = e a + b – 4 = 4 and b + a – 4 = 4 b = 8 – a ∈ Z Hence, 8 – a is the inverse of a ∈...
Answer: Consider, e be the identity element in I+ with respect to * such that a * e = a = e * a, ∀ a ∈ Q – {-1} a * e = a and e * a = a, ∀ a ∈ Q – {-1} a + e + ae = a and e + a + ea = a, ∀ a ∈ Q –...
Answer: Consider, e be the identity element in I+ with respect to * a * e = a = e * a, ∀ a ∈ I+ a * e = a and e * a = a, ∀ a ∈ I+ a + e = a and e + a = a, ∀ a ∈ I+ e = 0, ∀ a ∈ I+ Hence, 0 is the...
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.