Differentiate between:
Question: Differentiate between: (a) Myelinated and non-myelinated axons (b) Dendrites and axons (c) Rods and cones (d) Thalamus and Hypothalamus (e) Cerebrum and Cerebellum Solution: (a)Myelinated and non-myelinated axons (b)Dendrites and axons (c) Rods and cones (d)Thalamus and Hypothalamus (e) Cerebrum and Cerebellum...
Read More →Draw a triangle ABC with side BC = 7 cm,
Question: Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=7 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}, \angle \mathrm{A}=105^{\circ} .$ Then construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\triangle \mathrm{ABC}$. Solution: Steps of construction : 1. Construction $\triangle \mathrm{ABC}$ such that $\mathrm{BC}=7 \mathrm{~cm}$, $\angle \mathrm{ABC}=45^{\circ}$ and $\angle \mathrm{ACB}=30^{\circ}$ (i.e., $\angle \mathrm{BAC}=105^{\circ}$ ). 2. Draw any r...
Read More →Draw a triangle ABC with side BC = 6 cm,
Question: Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=6 \mathrm{~cm}, \mathrm{AB}=5 \mathrm{~cm}$ and $\angle \mathrm{ABC}=60^{\circ}$. Then construct a triangle whose sides are $\frac{\mathbf{3}}{\mathbf{4}}$ of the corresponding sides of the triangle $\mathrm{ABC}$. Solution: Steps of construction : 1. Draw a $\triangle \mathrm{ABC}$ with side $\mathrm{BC}=6 \mathrm{~cm}, \mathrm{AB}=5 \mathrm{~cm}$ and $\angle \mathrm{ABC}=60^{\circ}$. 2. Draw a ray $\mathrm{BX}$ making an acute ang...
Read More →Explain the following:
Question: Explain the following: (a) Role of Na+in the generation of action potential. (b) Mechanism of generation of light-induced impulse in the retina. (c) Mechanism through which a sound produces a nerve impulse in the inner ear. Solution: (a)Sodium ions play an important role in the generation of action potential. When a nerve fibre is stimulated, the membrane potential decreases. The membrane becomes more permeable to Na+ions than to K+ions. As a result, Na+diffuses from the outside to the...
Read More →Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Question: Calculate the total number of electrons present in 1.4 g of dinitrogen gas. Solution: Molar mass of dinitrogen (N2) = 28 g mol1 Thus, $1.4 \mathrm{~g}$ of $\mathrm{N}_{2}=\frac{1.4}{28}=0.05 \mathrm{~mol}$ $=0.05 \times 6.02 \times 10^{23}$ number of molecules $=3.01 \times 10^{23}$ number of molecules Now, 1 molecule of $\mathrm{N}_{2}$ contains 14 electrons. Therefore, $3.01 \times 10^{23}$ molecules of $N_{2}$ contains $=1.4 \times 3.01 \times 10^{23}$ = 4.214 1023electrons...
Read More →Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
Question: Calculate the total number of electrons present in 1.4 g of dinitrogen gas. Solution: Molar mass of dinitrogen (N2) = 28 g mol1 Thus, $1.4 \mathrm{~g}$ of $\mathrm{N}_{2}=\frac{1.4}{28}=0.05 \mathrm{~mol}$ $=0.05 \times 6.02 \times 10^{23}$ number of molecules $=3.01 \times 10^{23}$ number of molecules Now, 1 molecule of $\mathrm{N}_{2}$ contains 14 electrons. Therefore, $3.01 \times 10^{23}$ molecules of $N_{2}$ contains $=1.4 \times 3.01 \times 10^{23}$ = 4.214 1023electrons...
Read More →Answer briefly:
Question: Answer briefly: (a) How do you perceive the colour of an object? (b) Which part of our body helps us in maintaining the body balance? (c) How does the eye regulate the amount of light that falls on the retina? Solution: (a)Photoreceptors are cells that are sensitive to light. They are of two types rods and cones. These are present in the retina. Cones help in distinguishing colours. There are three types of cone cells those responding to green light, those responding to blue light, and...
Read More →Construct an isosceles triangle whose base is 8 cm and altitude 4 cm
Question: Construct an isosceles triangle whose base is $8 \mathrm{~cm}$ and altitude $4 \mathrm{~cm}$ and then another triangle whose sides are $\mathbf{1} \frac{\mathbf{1}}{\mathbf{2}}$ times the corresponding sides of the isosceles triangle. Solution: Steps of construction : 1. Construct isosceles $\triangle \mathrm{ABC}$ such that $\mathrm{BC}=8 \mathrm{~cm}$ and altitude $\mathrm{AD}=4 \mathrm{~cm}$. 2. Draw any ray $\mathrm{BX}$ making an acute angle with $\mathrm{BC}$ 3. Take $\mathrm{B}_...
Read More →Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C.
Question: Let A, B and C be the sets such that A B = A C and A B = A C. show that B = C. Solution: Let, A, B and C be the sets such thatand. To show: B = C Letx B $\Rightarrow x \in A \cup B \quad[B \subset A \cup B]$ $\Rightarrow x \in \mathrm{A} \cup \mathrm{C} \quad[\mathrm{A} \cup \mathrm{B}=\mathrm{A} \cup \mathrm{C}]$ $\Rightarrow x \in \mathrm{A}$ or $x \in \mathrm{C}$ Case I x A Also,x B $\therefore x \in A \cap B$ $\Rightarrow x \in \mathrm{A} \cap \mathrm{C} \quad[\because \mathrm{A} \...
Read More →Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
Question: Calculate the temperature of 4.0 mol of a gas occupying 5 dm3at 3.32 bar. $\left(\mathrm{R}=0.083\right.$ bar $\left.\mathrm{dm}^{3} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ Solution: Given, n= 4.0 mol V= 5 dm3 p= 3.32 bar R = 0.083 bar dm3K1mol1 The temperature (T) can be calculated using the ideal gas equation as: $p V=n \mathrm{R} T$ $\Rightarrow T=\frac{p V}{n \mathrm{R}}$ $=\frac{3.32 \times 5}{4 \times 0.083}$ $=50 \mathrm{~K}$ Hence, the required temperature is 50 K....
Read More →Give a brief account of:
Question: Give a brief account of: (a) Mechanism of synaptic transmission (b) Mechanism of vision (c) Mechanism of hearing Solution: (a) Mechanism of synaptic transmission Synapse is a junction between two neurons. It is present between the axon terminal of one neuron and the dendrite of next neuron separated by a cleft. There are two ways of synaptic transmission. (1)Chemical transmission (2)Electrical transmission 1.Chemical transmission When a nerve impulse reaches the end plate of axon, it r...
Read More →Construct a triangle with sides 5 cm,
Question: Construct a triangle with sides $5 \mathrm{~cm}, 6 \mathrm{~cm}$ and $7 \mathrm{~cm}$ and then another triangle whose sides are$\frac{\mathbf{7}}{\mathbf{5}}$ of the corresponding sides of the first triangle. Solution: Steps of Construction : 1. Construct a $\triangle \mathrm{ABC}$ such that $\mathrm{AB}=5 \mathrm{~cm}$, $\mathrm{BC}=7 \mathrm{~cm}$ and $\mathrm{AC}=6 \mathrm{~cm}$ 2. Draw a ray $\mathrm{BX}$ such that $\angle \mathrm{CBX}$ is an acute angle. 3. Mark 7 points $X_{1}, X...
Read More →In each of the following, determine whether the statement is true or false. If it is true,
Question: In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i)Ifx A and A B, thenx B (ii)If A B and B C, then A C (iii)If A B and B C, then A C (iv)If A B and B C, then A C (v)Ifx A and A B, thenx B (vi)If A B andx B, thenx A Solution: (i)False Let A = {1, 2} and B = {1, {1, 2}, {3}} Now,$2 \in\{1,2\}$ and $\{1,2\} \in\{\{3\}, 1,\{1,2\}\}$ $\therefore A \in B$ However, $2 \notin\{\{3\}, 1,\{1,2\}\}$ (ii)False Le...
Read More →A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame.
Question: A student forgot to add the reaction mixture to the round bottomed flask at 27 C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 C. What fraction of air would have been expelled out? Solution: Let the volume of the round bottomed flask beV. Then, the volume of air inside the flask at 27 C isV. Now, V1=V T1= 27C = 300 K V2=? T2= 477 C = 750 K According to Charless law, $...
Read More →A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead he/she placed the flask on the flame.
Question: A student forgot to add the reaction mixture to the round bottomed flask at 27 C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477 C. What fraction of air would have been expelled out? Solution: Let the volume of the round bottomed flask beV. Then, the volume of air inside the flask at 27 C isV. Now, V1=V T1= 27C = 300 K V2=? T2= 477 C = 750 K According to Charless law, $...
Read More →Write short notes on the following:
Question: Write short notes on the following: (a) Neural coordination (b) Forebrain (c) Midbrain (d) Hindbrain (e) Retina (f) Ear ossicles (g) Cochlea (h) Organ of Corti (i) Synapse Solution: (a)Neural coordination The neural system provides rapid coordination among the organs of the body. This coordination is in the form of electric impulses and is quick and short lived. All the physiological processes in the body are closed linked and dependent upon each other. For example, during exercise, ou...
Read More →Construct a triangle of sides 4 cm,
Question: Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the correspoding sides of the first triangle. Solution: Steps of construction : 1. Construct ABC, such that AB = 4 cm, BC = 5 cm and CA = 6 cm. 2. Draw any ray BX making an acute angle with BC (below the side BC). 3. Mark three points $\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}$ on $\mathrm{BX}$ such that $\mathrm{BB}_{1}=\mathrm{B}_{1} \mathrm{~B}_{2}=\mathrm{B}_{2} \mathr...
Read More →34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure.
Question: 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 C and 0.1 bar pressure. What is the molar mass of phosphorus? Solution: Given, p= 0.1 bar V= 34.05 mL = 34.05 103L = 34.05 103dm3 R = 0.083 bar dm3K1mol1 T= 546C = (546 + 273) K = 819 K The number of moles (n) can be calculated using the ideal gas equation as: $p V=n \mathrm{R} T$ $\Rightarrow n=\frac{p V}{R T}$ $=\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819}$ $=5.01 \times 10^{-5} \mathrm{~mol}$ Therefore, molar mass of p...
Read More →Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8.
Question: Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. Solution: Steps of construction : 1. Draw AB = 7.6 cm. 2. AX is any ray making an acute angle with AB above the line AB. 3. Draw ray $B Y$ below the line $A B$ and parallel to the ray $A X$ by constructing $\angle A B Y=\angle B A X$. 4. Mark points $A_{1}, A_{2}, \ldots, A_{5}$ on $A X$ and $B_{1}, B_{2}$, .....,$\mathrm{B}_{8}$ on $\mathrm{BY}$ such that $\mathrm{AA}_{1}=\mathrm{A}_{1} \math...
Read More →Decide, among the following sets, which sets are subsets of one and another:
Question: Decide, among the following sets, which sets are subsets of one and another: $A=\left\{x: x \in R\right.$ and $x$ satisfy $\left.x^{2}-8 x+12=0\right\}$ $B=\{2,4,6\}, C=\{2,4,6,8 \ldots\}, D=\{6\}$ Solution: $A=\left\{x: x \in R\right.$ and $x$ satisfies $\left.x^{2}-8 x+12=0\right\}$ 2 and 6 are the only solutions of $x^{2}-8 x+12=0$ $\therefore A=\{2,6\}$ $B=\{2,4,6\}, C=\{2,4,6,8 \ldots\}, D=\{6\}$ $\therefore D \subset A \subset B \subset C$ Hence, $A \subset B, A \subset C, B \sub...
Read More →Draw labelled diagrams of the following:
Question: Draw labelled diagrams of the following: (a) Neuron (b) Brain (c) Eye (d) Ear Solution: (a)Neuron (b)Brain (c)Eye (d)Ear...
Read More →Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure.
Question: Density of a gas is found to be $5.46 \mathrm{~g} / \mathrm{dm}^{3}$ at $27^{\circ} \mathrm{C}$ at 2 bar pressure. What will be its density at STP? Solution: Given, The density (d2) of the gas at STP can be calculated using the equation, Hence, the density of the gas at STP will be 3 g dm3....
Read More →In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French.
Question: In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages? Solution: Let F be the set of people in the committee who speak French, and S be the set of people in the committee who speak Spanish n(F) = 50,n(S) = 20,n(SF) = 10 We know that: n(SF) =n(S) +n(F) n(SF) = 20 + 50 10 = 70 10 = 60 Thus, 60 people in the committee speak at least one of the two languages....
Read More →In a group of 65 people, 40 like cricket, 10 like both cricket and tennis.
Question: In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis? Solution: Let C denote the set of people who like cricket, and T denote the set of people who like tennis n(CT) = 65,n(C) = 40,n(CT) = 10 We know that: n(CT) =n(C) +n(T) n(CT) 65 = 40 +n(T) 10 ⇒65 = 30 +n(T) ⇒n(T) = 65 30 = 35 Therefore, 35 people like tennis. Now, (T C)(TC) = T Also, (T C)(TC) =Φ n(T) =n(T C) +n(TC) ⇒35 =n(T C) + 10 ⇒n(T C) = 35 1...
Read More →In a group of 70 people, 37 like coffee, 52 like tea,
Question: In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea? Solution: Let C denote the set of people who like coffee, and T denote the set of people who like tea n(CT) = 70,n(C) = 37,n(T) = 52 We know that: n(CT) =n(C) +n(T) n(CT) 70 = 37 + 52 n(CT) ⇒70 = 89 n(CT) ⇒n(CT) = 89 70 = 19 Thus, 19 people like both coffee and tea....
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