Mehapatel

सदाबहार वन
पर्णपाती वन

(i) यह वन उन क्षेत्रों में पाए जाते हैं जहाँ 200 सेमी. से अधिक वर्षा होती है।(ii) इनके वृक्षों में पतझड़ होने का कोई समय निश्चित नहीं होता अतः यह वन साल भर हरे भरे रहते हैं।(iii) इन वनों में पाए जाने वाले व्यापारिक महत्व के वृक्ष आबनूस, रोसवुड, रबड़ आदि हैं।


(i) यह वन उन क्षेत्रों में पाए जाते हैं जहां 70 सेमी. से 200 सेमी. तक वर्षा होती है।(ii) इस प्रकार के वनों के वृक्ष शुष्क ग्रीष्म ऋतु में लगभग छः से आठ सप्ताह के लिए अपनी पत्तियाँ गिरा देते हैं।(iii) इन वनों में पाए जाने व्यापारिक महत्व के वृक्ष बाँस, साल, शहतूत आदि हैं।

Laboratory preparation of methane: It is prepared in the laboratory by heating a mixture of sodium acetate with soda lime.
Experiment: Powdered sodium acetate is mixed with four times the amount of soda lime. The mixture is taken in a hard glass tube. It is fitted with the delivery tube.As the contents of the tube are heated, methane gas is produced. It is collected by the downward displacement of water.

a) When an object is placed in front of the lens between its optical centre and principal focus, the image is formed beyond 2F1 (on the same side of the object), and the ray diagram is obtained is as follows:b) The object distance (u) and the image distance (v) are as shown below. Since both the image and the object lie in a direction opposite to the direction of incoming rays, the magnitude will be negative for both.

 
The relation between u, v and f is given by the lens formula,  c) Given, u = -20 cm m = -1 Since magnification  is given as, Focal length (f) can be calculated as: Thus, t

The velocity does not depends on the mass of the particle because the acceleration due to gravity under which the particle falls does not depend on mass.

So,    BP = BQ(Tangents from external point B)But    BP = 27 cm⇒    BQ = 27 cmIt is given that BC = 38 cm⇒    BQ + CQ = 38⇒    27 + CQ = 38⇒    CQ = 11 cm⇒ CQ = CR (Tagents from an external point C)But    CQ = 11 cm⇒    CR = 11 cm
It is given that : CD = 25 cm⇒    CR + DR = 25⇒    11 + DR = 25⇒    DR = 14 cmSince, tangent to a circle is perpendicular to the radius through the point of contact.∴ ∠ORD = ∠OSD = 90°It is given that∠D = 90°Now, in quadrilateral ORDS,∠ORD = ∠OSD = ∠RDS = ∠ROS = 90°and    OR = OS [radii of circle]Therefore, ORDS is a squareSo,    OR = DR = 14 cmHence    r = 14 cm.

Maharashtra Board NCERT solutions for Class 11.Get free online study material for Class 11.Find NCERT solutions textbook questions and answers and download NCERT books from Zigya Book-Store.

A sphere of radius r is divided into 4 identical parts∴  Radius of each part  = r unitsEach part has one curved surface and 2 semicircular faces∴  T.S.A. of each part  = 
                                ∴ T.S.A. of four parts 
                               
       

(i) Be2 molecule: The electronic configuration of Be(Z = 4) is:4 Be 1s2 2s1Be2 molecule is formed by the overlap of atomic orbitals of both beryllium atoms.Number of valence electrons in Be atom = 2Thus in the formation of Be2 molecule, two outer electrons of each Be atom i.e. 4 in all, have to be accommodated in various molecular orbitals in the increasing order of their energies..The molecular orbital electronic configuration,Magnetic property: Since bond order is zero, Be2 molecule does not exist. It is diamagnetic due to the absence of any unpaired electron.B2 molecule: The electronic con

Total number of all possible outcomes = 52.i,e,. n(S)=52(i) Let B be the favourable outcomes of getting red,thenn (B) = 26.Therefore, P(B) = 
(ii) Let C be the favourable outcomes of getting a card either red or king, then n(C)= 28.
Therefore,  P(C) = (iii) Let D be the favourable outcomes of getting red and a king, then n(D)=2 Therefore,P(D) = 
(iv) Let E be the favourable outcomes of getting a red face card, thenn(E) = 6Therefore,  P(E) = 
(v) Let F be the favourable outcomes of getting 2 of spades, thenE = {1}⇒    n(E) = 1Therefore,  P(E) = 
(vi) Let G be the favourable outcomes of ge

h = 1 m = 100 cm2r = 140 cmTotal surface area of the closed cylindrical tank = 2πr(h + r)
Hence, 7.48 square metres of the sheet are required.

According to question, Required distance  =                                 

 NiCl42-, there is Ni2+ ion, However, in presence of weak field Cl- ligands, NO pairing of d-electrons occurs. Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl- ligands in tetrahedral geometry. As there are unpaired electrons in the d-orbitals, NiCl42- is paramagnetic. Since it have two unpaired electron electron therefore the magnetic moment :Magnetic moment = n(n+2)                         = 2(2+2)                        =8                         =2.82
In [Ni(CN)4]2-, there is Ni2+ ion for which the electronic configuration in the valence shell is 3d8 4s0.
 In presence o

We have    12x =                    36 =∴        12x + 36 = [(223)x] + (223) 3                         =

Given, length of the elctric dipole = 4cm=4×10-2 mangle made with the elctric field-θ=60∘Torque experienced by the dipole-τ = 43 Nmcharge on the dipole-q=±8nC = ±8×10-9 Ci) magnitude of Electric field-E                          τ= pE sinθ    = q × 2a × E sinθ     ⇒  43 = 8×10-9 × 4×10-2 × E × sin 60°    ⇒       E = 43×232 ×10-11×3    = 14×1011     = 2.5 × 1010 NC-1(ii) magnitude of potential energy of the dipole  U = -pE cos θ    = -q×2a×E cos θ    =-8×10-9×2×10-2×2.5×1010 cos 60°= - 2J.

The consumer will buy 75 (= 50 + 25) units.

Lateral displacement of an emergent ray depends on: (i) Angle of incidence,(ii) Thickness of the glass slab, and (iii) Refractive index of the slab material.

Glycerol with high boiling point (290oC) can be separated from spent lye by distillation under reduced pressure. This process is used to purify liquids having very high boiling points. By this process, the liquid is made to boil at a lower temperature than its boiling point by lowering the pressure on its surface.

पुरातत्वविदों को विजय नगर की खुदाई के दौरान विभिन्न प्रकार के अवशेष मिले। शहर के केंद्र में कुछ 52 बड़े भव्य तथा अन्य स्थानों की तुलना में अधिक सुरक्षात्मक ढंग से बनाए गए थे। इनकी बनावट को देखकर यह अनुमान लगाया जा सकता है कि किसी स्थल से विजयनगर साम्राज्य का प्रशासन चलता था इसलिए पुरातत्वविदों ने इसे शाही या राजकीय केंद्र कहां है।
इस क्षेत्र में शासक का आवास, 60 से अधिक मंदिर तथा बड़े-बड़े सभा स्थल है। शहर के इसी हिस्से में महानवमी डिब्बा जैसे भव्य चबूतरा हैं जहाँ शासक अपनी भव्यता, शक्ति का प्रदर्शन करते थे। इस तरह के शासकों का आयोजन स्थल भी यही स्वीकारा जाता है। विभिन्न मंडपों से आंगन भी

पहले दस तत्व हैं- H, He, Li, Be, B, C, N, O, F तथा Neइन सब तत्वों में से धातु हैं- Li तथा BeB एक मिश्रधातु है।

High Powered Money. High powered money or monetary base refers to the money produced by R.B.I. and Government of India. Alternatively total liability of monetary authority of the country and R.B.I. is called monetary base or high powered money (H). It consists of (i) currency (notes and coins) in the hands of public (C), (ii) Cash reserve of commercial banks (R) and (iii) Other deposits with R.B.I. (OD). Symbolically:H = C + R + ODHigh powered money is different from ordinary money (M1) which consists of (i) currency held by public (C), (ii) Demand deposits in banks (DD), and (iii) Other depo